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>
using namespace std;
int edge[510][510], vis[510], con, N, M, f, t;
void dfs(int x) {
  vis[x] = 1;
  for (int i = 1; i <= N; i++)
    if (edge[x][i]) {
      if (!vis[i])
        dfs(i);
      else if (vis[i] == 1)
        con++;
      if (con >= 2) break;
    }
  vis[x] = 2;
}
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 1; i <= M; i++) {
    scanf("%d%d", &f, &t);
    edge[f][t] = 1;
  }
  for (int i = 1; i <= N; i++) {
    con = 0;
    memset(vis, 0, sizeof vis);
    dfs(i);
    for (int j = 1; j <= N; j++)
      if (j != i) {
        if (!vis[j]) dfs(j);
        if (con >= 2) break;
      }
    if (con <= 1) return puts("YES"), 0;
  }
  return puts("NO"), 0;
}
