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, in[505], tmp[505];
bool vis[505];
vector<int> g[505];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    ++in[b];
    g[a].push_back(b);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      tmp[j] = in[j];
      vis[j] = 0;
    }
    tmp[i] = max(0, tmp[i] - 1);
    queue<int> cur;
    for (int j = 1; j <= n; j++) {
      if (tmp[j] <= 0) {
        cur.push(j);
        vis[j] = 1;
      }
    }
    while (!cur.empty()) {
      int u = cur.front();
      cur.pop();
      for (int i = 0; i < g[u].size(); i++) {
        int v = g[u][i];
        if (tmp[v] == 0) continue;
        if (--tmp[v] == 0) {
          cur.push(v);
          vis[v] = 1;
        }
      }
    }
    bool flag = 1;
    for (int i = 1; i <= n; i++) {
      if (tmp[i] != 0) {
        flag = 0;
        break;
      }
    }
    if (flag) {
      cout << "YES" << endl;
      return 0;
    }
  }
  cout << "NO" << endl;
}
