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;
const int MOD = 1e9 + 7;
const int qq = 500 + 10;
const int INF = 1e9 + 10;
vector<int> G[qq];
bool vis[qq];
int n, m, dag[qq], tmp[qq];
void dfs(int u) {
  vis[u] = true;
  for (int i = 0; i < G[u].size(); ++i) {
    int v = G[u][i];
    dag[v]--;
    if (!vis[v]) {
      if (dag[v] <= 0) {
        dfs(v);
      }
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int a, b, i = 0; i < m; ++i) {
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    tmp[b]++;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      dag[j] = tmp[j];
    }
    dag[i]--;
    memset(vis, false, sizeof vis);
    for (int j = 1; j <= n; ++j) {
      if (!vis[j] && dag[j] <= 0) {
        dfs(j);
      }
    }
    bool ok = true;
    for (int j = 1; j <= n; ++j) {
      if (!vis[j]) {
        ok = false;
      }
    }
    if (ok) {
      printf("YES\n");
      return 0;
    }
  }
  printf("NO\n");
  return 0;
}
