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#include <bits/stdc++.h>
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using namespace std;
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typedef pair<int, int> PII;
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const int N = 512 * 3 + 10;
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char a[N][N];//疑问：第一维为什么也要开N这么大，按理说开N/2就够了
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PII b[N][N];
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int width;
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//放过位置0，后面是10个数字
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//此数组的含义：穷举出level=1，2，3，4，... ，10时(从下往上)，子树开始的位置。
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const int col[11] = {0, 1, 3, 6, 12, 24, 48, 96, 192, 384, 768};
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/**
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* 功能：自顶向下画二叉树
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* @param r 从哪一行开始
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* @param c 从哪一列开始
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* @param depth 深度
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*/
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void dfs(int r, int c, int depth) {
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//此位置放上o
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a[r][c] = 'o';
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//如果到了叶子结点，递归结束
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if (depth == 1) return;//这个depth是用来控制递归终止的，判断是不是到了叶子
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//树枝的个数,从当前结点到下一层结点间树枝的个数是多少,可以转化为求相邻结点的列差值
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int cnt = col[depth] - col[depth - 1] - 1;
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//下一步的左儿子，右儿子的位置
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int lSonRow = r + 1, lSonCol = c - 1, rSonRow = r + 1, rSonCol = c + 1;
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//执行几次，画出树干
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for (int i = 1; i <= cnt; i++) {
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a[lSonRow++][lSonCol--] = '/'; //在循环中动态修改左右儿子的位置坐标，继续填充
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a[rSonRow++][rSonCol++] = '\\';
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}
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//递归填充左子树
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dfs(lSonRow, lSonCol, depth - 1);
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//递归填充右子树
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dfs(rSonRow, rSonCol, depth - 1);
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}
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/**
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* 功能：输出二维数组
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*/
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void print() {
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//width为最长有效数据范围
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for (int i = 1; i <= width; i++) {
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for (int j = 1; j <= width; j++)
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cout << a[i][j];
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cout << endl;
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}
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}
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/**
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* 递归删除结点
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* @param r 行
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* @param c 列
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*/
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void delNode(int r, int c) {
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a[r][c] = ' ';
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if (a[r - 1][c - 1] == '\\') delNode(r - 1, c - 1);//向左上进行探索删除,向上不能删除o
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else if (a[r - 1][c + 1] == '/') delNode(r - 1, c + 1);//向右上进行探索删除,向上不能删除o
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//上面的可以使用else,因为一个位置只能向上有一条路径，不能既有左上，又有右上
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//向下就很暴力了，不管是/ \ o全部干掉
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if (a[r + 1][c - 1] == '/' || a[r + 1][c - 1] == 'o') delNode(r + 1, c - 1);
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if (a[r + 1][c + 1] == '\\' || a[r + 1][c + 1] == 'o') delNode(r + 1, c + 1);
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}
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//生成每一行每一列每一个o的具体位置
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//因为题目给的数据是第几行第几列的相对位置，需要转换为在数组中的绝对位置
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void fillReleation() {
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int row = 1;
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for (int i = 1; i <= width; i++) {
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int cnt = 0;
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bool found = false;
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for (int j = 1; j <= width; j++) {
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if (a[i][j] == 'o') {//本行中找到o, cnt记录个数
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cnt++;
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found = true; //标识为找到，本行找到后，row++,找不到，不用++
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b[row][cnt] = {i, j};
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}
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}
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if (found) row++;
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}
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}
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int main() {
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int m; //绘制的二叉树层数
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int delCnt; //要删除结点的个数
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cin >> m >> delCnt;
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//行和列都设置为空格
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memset(a, ' ', sizeof a);
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//给定m情况下的底层宽度
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width = pow(2, m - 1) * 3;
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//填充
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dfs(1, col[m], m);
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//生成具体位置对应关系表
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fillReleation();
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//删除结点
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int x, y;
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for (int i = 1; i <= delCnt; i++) {
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cin >> x >> y;
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delNode(b[x][y].first, b[x][y].second);
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}
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//打印
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print();
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return 0;
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}
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