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;
vector<int> p;
vector<int> cor;
int total;
void dfs_visit(vector<vector<int>>& adj, int i) {
  cor[i] = 1;
  for (int j : adj[i]) {
    if (cor[j] == 1) {
      total++;
    }
    if (cor[j] == 0) {
      p[j] = i;
      dfs_visit(adj, j);
    }
  }
  cor[i] = 2;
}
int dfs(vector<vector<int>>& adj, int n) {
  int ans = 10;
  for (int i = 1; i <= n; i++) {
    total = 0;
    p.assign(n + 1, 1);
    cor.assign(n + 1, 0);
    total = 0;
    int j = i;
    if (cor[j] == 0) {
      dfs_visit(adj, j);
    }
    for (int j = 1; j <= n; j++) {
      if (cor[j] == 0) {
        dfs_visit(adj, j);
      }
    }
    ans = min(ans, total);
  }
  return ans;
}
int main() {
  int n, m, a, b;
  cin >> n >> m;
  vector<int> vetor;
  vetor.assign(n + 1, 0);
  vector<vector<int>> adj;
  vector<int> lista;
  lista.clear();
  adj.assign(n + 1, lista);
  for (int i = 1; i <= m; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
  }
  int ans = dfs(adj, n);
  if (ans <= 1)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}
