You are given a [directed graph](https://en.wikipedia.org/wiki/Directed_graph) consisting of n vertices and m edges (each edge is directed, so it can be traversed in only one direction). You are allowed to remove at most one edge from it.

Can you make this graph [acyclic](https://en.wikipedia.org/wiki/Directed_acyclic_graph) by removing at most one edge from it? A directed graph is called acyclic iff it doesn't contain any cycle (a non-empty path that starts and ends in the same vertex).

Input

The first line contains two integers n and m (2 ≤ n ≤ 500, 1 ≤ m ≤ min(n(n - 1), 100000)) — the number of vertices and the number of edges, respectively.

Then m lines follow. Each line contains two integers u and v denoting a directed edge going from vertex u to vertex v (1 ≤ u, v ≤ n, u ≠ v). Each ordered pair (u, v) is listed at most once (there is at most one directed edge from u to v).

Output

If it is possible to make this graph acyclic by removing at most one edge, print YES. Otherwise, print NO.

Examples

Input

3 4
1 2
2 3
3 2
3 1


Output

YES


Input

5 6
1 2
2 3
3 2
3 1
2 1
4 5


Output

NO

Note

In the first example you can remove edge <image>, and the graph becomes acyclic.

In the second example you have to remove at least two edges (for example, <image> and <image>) in order to make the graph acyclic.
#include <bits/stdc++.h>
int n, m, ln;
int a[1000005], tl[1000005], inde[10005], inde2[10005], nx[1000005];
int q[1000005], tail, head = 1;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) tl[i] = ++ln;
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[tl[x] = nx[tl[x]] = ++ln] = y;
    inde[y]++;
  }
  for (int i = 1; i <= n; i++) {
    if (inde[i]) {
      for (int j = 1; j <= n; j++) inde2[j] = inde[j];
      inde2[i]--;
      head = 1, tail = 0;
      for (int j = 1; j <= n; j++)
        if (inde2[j] == 0) q[++tail] = j;
      while (head <= tail) {
        int x = q[head++];
        for (int j = nx[x]; j; j = nx[j]) {
          inde2[a[j]]--;
          if (inde2[a[j]] == 0) q[++tail] = a[j];
        }
      }
      bool flg = true;
      for (int j = 1; j <= n; j++)
        if (inde2[j] > 0) flg = false;
      if (flg) return !printf("YES");
    }
  }
  printf("NO");
  return 0;
}
