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;
using ll = long long;
const int N = 1E5;
int a[N + 7];
vector<int> mp[N];
int in[N];
int in1[N];
bool vis[N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    mp[x].push_back(y);
    in[y]++;
  }
  memcpy(in1, in, sizeof(in1));
  int flag = 1;
  for (int i = 1; i <= n; i++) {
    if (in[i] == 0) continue;
    in[i]--;
    queue<int> q;
    for (int j = 1; j <= n; j++) {
      if (in[j] == 0) {
        q.push(j);
      }
    }
    int cnt = 0;
    while (!q.empty()) {
      cnt++;
      int u = q.front();
      q.pop();
      int len = mp[u].size();
      for (int j = 0; j < len; j++) {
        int v = mp[u][j];
        in[v]--;
        if (!in[v]) q.push(v);
      }
    }
    if (n == cnt) {
      cout << "YES" << endl;
      return 0;
    }
    memcpy(in, in1, sizeof(in));
  }
  cout << "NO" << endl;
}
