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#include <bits/stdc++.h>
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using namespace std;
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const int N = 30; //26个结点是极限
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const int M = 610; //600条边是极限
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int in[N]; //入度数组
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int tmp[N]; //入度临时操作数组
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int n; //表示需要排序的元素数量
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int m; //表示将给出的形如 A<B 的关系的数量
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bool st[N]; //是否使用过
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//实在想不出好主意了,用数字转为字符，再拼接到字符串，现在看来最好的办法就是写成一个常量字符串，然后substr去截取
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string chr = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
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struct Node {
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int n;
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string path;
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};
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//链式前向星
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int idx;
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int head[N]; //链表头数组
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struct Edge {
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int to, next;
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} edge[M];
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//从u向v连一条边，本题无权值概念,头插法
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void add(int u, int v) {
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edge[++idx].to = v;
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edge[idx].next = head[u];
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head[u] = idx;
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}
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//拓扑排序，检查是否存在环，有环就是存在矛盾
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bool topSort() {
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//清空状态数组
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memset(st, 0, sizeof st);
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queue<int> q; //拓扑排序用的队列
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int sum = 0;//入队列数量
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//判断现在是不是有环，有环则退出。入度为0的结点入队列，进行拓扑排序
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for (int j = 1; j <= n; j++) if (!tmp[j]) q.push(j);
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//拓扑排序
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while (!q.empty()) {
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int u = q.front();
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q.pop();
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sum++;
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//不是环中结点
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st[u] = true;
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//遍历每条出边
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for (int j = head[u]; j; j = edge[j].next) {
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int y = edge[j].to;
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if (!--tmp[y]) q.push(y);//在删除掉当前结点带来的入度后，是不是入度为0了，如果是将点y入队列
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}
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}
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//存在没进过队列的结点，存在表示有环。
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return sum < n;
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}
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int main() {
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//读入结点数和边数
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cin >> n >> m;
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//读入关系，建图，边建图边判断
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for (int i = 1; i <= m; i++) {
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string input;
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cin >> input;
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char x = input[0], y = input[2];//input[1]='<' 这个是死的，没用
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//建图
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int a = x - 'A' + 1, b = y - 'A' + 1;
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add(a, b); //建立单边有向图,描述a<b
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in[b]++; //入度数组++
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/*****下面的代码是拓扑排序的内容*******************************************************/
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//入度数组复制出来一份,因为拓扑排序需要不断的更改入度数组值，后面还需要原始的入度数组，所以需要复制出来一份。
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memcpy(tmp, in, sizeof in);
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bool existCircle = topSort();
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if (existCircle) {
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printf("Inconsistency found after %d relations.", i);
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exit(0);
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}
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/*****拓扑排序结束*********************************************************************/
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//到这里应该是无环
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//从入度为0的点出发，bfs看看是不是能走完n步，走完说明可以确定所有结点的大小关系，确定了就结束
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int cnt = 0;
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int startNode;
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for (int j = 1; j <= n; j++) if (!in[j]) startNode = j, cnt++;
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if (cnt > 1) continue; //存在两个及以上入度为零的点，是无法确定先后顺序的,需要继续读入关系数据
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//从startNode出发，看看能不能走出去n轮，如果能，就是可以确定所有结点的顺序，如果不能，就需要继续录入关系才能确定
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string path;
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queue<Node> q;
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q.push({startNode, chr.substr(startNode - 1, 1)});
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//清空状态数组，准备再战!
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memset(st, 0, sizeof st);
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while (!q.empty()) {
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Node u = q.front();
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q.pop();
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st[u.n] = true;
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if (u.path.size() > path.size())path = u.path;
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//遍历每条出边
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for (int j = head[u.n]; j; j = edge[j].next) {
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int y = edge[j].to;
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q.push({y, u.path + chr.substr(y - 1, 1)});
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}
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}
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//如果还存在没访问到的结点，那么就是无法完成排序,需要继续录入关系
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if (path.size() < n) continue;
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//输出路径
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printf("Sorted sequence determined after %d relations: %s.", i, path.c_str());
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exit(0);
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}
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//到达这里说明最终还没有确定下来所有结点关系，提示无法确定。
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printf("Sorted sequence cannot be determined.");
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return 0;
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}
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