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 n, m, a[502][502], ok, x, y;
bitset<505> v, ins;
pair<int, int> q[100005];
void dfs(int x) {
  v[x] = 1;
  ins[x] = 1;
  for (int i = 1; i <= n; ++i) {
    if (a[x][i] && !v[i]) {
      dfs(i);
      continue;
    }
    if (a[x][i] && ins[i]) {
      ok++;
    }
  }
  ins[x] = 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> x >> y, a[x][y] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    ok = 0;
    v = 0;
    dfs(i);
    for (int j = 1; j <= n; ++j)
      if (!v[j]) dfs(j);
    if (ok < 2) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
  return 0;
}
