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#include <bits/stdc++.h>
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using namespace std;
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int n, m;
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/*
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测试用例：
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4 3
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1 2
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1 3
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1 4
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样例输出：
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0
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1
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2
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3
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*/
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const int N = 110;
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struct Node {
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int l, r, f;
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} tr[N];
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int root[N], rs;
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bool a[N][N];
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int main() {
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//文件输入
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freopen("DuoChaShuToErChaShu.in", "r", stdin);
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scanf("%d %d", &n, &m);
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for (int i = 1; i <= m; i++) { // m条边
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int x, y;
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scanf("%d %d", &x, &y);
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a[x][y] = true; // x->y有一条边，先保存起来
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}
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for (int i = 1; i <= n; i++) //双层循环，枚举每个点到点的关系
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for (int j = 1; j <= n; j++)
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if (a[i][j]) { //如果i->j有一条边
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if (tr[i].l == 0) { // i没有左儿子
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tr[i].l = j; // j就是i的左儿子
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tr[j].f = i; // i是j的父亲
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} else {
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int t = tr[i].l; //找到i的左儿子
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while (tr[t].r) t = tr[t].r; //一路向下找i的右儿子，右儿子,右儿子...
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tr[t].r = j; // j做为最终的右儿子
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tr[j].f = t; // j的父亲是最终的右儿子
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}
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}
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//处理森林
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for (int i = 1; i <= n; i++)
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if (!tr[i].f) root[++rs] = i; //将每个子树的根加入到root数组
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for (int i = rs; i > 1; i--) { //保留一个根，其它的根合并到前一个根中，从右向左当做右儿子加入
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tr[root[i - 1]].r = root[i];
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tr[root[i]].f = root[i - 1];
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}
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//输出每个点的父节点号
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for (int i = 1; i <= n; i++) printf("%d\n", tr[i].f);
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return 0;
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}
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