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;
int g[505][505] = {0};
int vis[505] = {0};
int ring[505] = {0};
int nring = 0;
int chk[505] = {0};
int checkring(int cur) {
  if (chk[cur] == -1) return 0;
  chk[cur] = -1;
  for (int next = 1; next <= n; ++next) {
    if (chk[next] != 1 && g[cur][next] == 1) {
      if (0 == checkring(next)) {
        return 0;
      }
    }
  }
  chk[cur] = 1;
  return 1;
}
int findring(int cur, int rdx) {
  if (vis[cur] == -1) {
    ring[rdx] = cur;
    nring = rdx;
    return 1;
  }
  vis[cur] = -1;
  for (int next = 1; next <= n; ++next) {
    if (vis[next] != 1 && g[cur][next] == 1) {
      if (1 == findring(next, rdx + 1)) {
        ring[rdx] = cur;
        return 1;
      }
    }
  }
  vis[cur] = 1;
  return -1;
}
int main() {
  scanf("%d %d", &n, &m);
  int a1, a2;
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &a1, &a2);
    g[a1][a2] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    nring = 0;
    if (1 == findring(i, 0)) {
      break;
    }
  }
  int pflag = 0, rflag;
  if (nring == 0) {
    printf("YES");
    return 0;
  }
  ++nring;
  for (int i = 0; i < nring && pflag == 0; ++i) {
    memset(chk, 0, sizeof chk);
    g[ring[i]][ring[(i + 1) % nring]] = 0;
    rflag = 1;
    for (int t = 1; t <= n && rflag; ++t) rflag = checkring(t);
    pflag = rflag;
    g[ring[i]][ring[(i + 1) % nring]] = 1;
  }
  if (pflag)
    printf("YES");
  else
    printf("NO");
}
