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> graf[505];
int in[505], cop[505];
bool don[505];
int n, m, f;
queue<int> q;
int del(int v) {
  int d = 0;
  q.push(v);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    don[u] = true;
    d++;
    for (int i = 0; i < graf[u].size(); i++) {
      in[graf[u][i]]--;
      if (in[graf[u][i]] == 0 && graf[u][i] != v) q.push(graf[u][i]);
    }
  }
  return d;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    in[b]++;
    graf[a].push_back(b);
  }
  int x;
  for (int i = 1; i <= n; i++)
    if (in[i] == 0 && !don[i]) x = del(i);
  for (int i = 1; i <= n; i++)
    if (in[i] != 0) f++;
  if (f == 0) {
    cout << "YES" << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) cop[j] = in[j];
    if (in[i] == 1 && del(i) == f) {
      cout << "YES" << endl;
      return 0;
    }
    for (int j = 1; j <= n; j++) in[j] = cop[j];
  }
  cout << "NO" << endl;
  return 0;
}
