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<long long int> G[505];
long long int n, m, inDegree[505], vis[505], totalREM = 0, IN[505];
long long int backEdge(long long int source) {
  long long int res = 0;
  queue<long long int> q;
  q.push(source);
  while (!q.empty()) {
    long long int u = q.front();
    q.pop();
    res++;
    for (int i = 0; i < G[u].size(); i++) {
      long long int v = G[u][i];
      if (v != source) {
        IN[v]--;
        if (IN[v] == 0) q.push(v);
      }
    }
  }
  return res;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    long long int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    inDegree[b]++;
  }
  queue<long long int> q;
  for (int i = 1; i <= n; i++) {
    if (inDegree[i] == 0) q.push(i), vis[i] = 1;
  }
  long long int cnt = 0;
  while (!q.empty()) {
    long long int u = q.front();
    q.pop();
    cnt++;
    for (int i = 0; i < G[u].size(); i++) {
      long long int v = G[u][i];
      inDegree[v]--;
      if (inDegree[v] == 0) q.push(v), vis[v] = 1;
    }
  }
  totalREM = n - cnt;
  if (totalREM == 0) {
    cout << "YES" << endl;
    return 0;
  }
  bool solution = false;
  for (int i = 1; i <= n; i++) {
    if (inDegree[i] == 1) {
      for (int i = 1; i <= n; i++) IN[i] = inDegree[i];
      if (backEdge(i) == totalREM) {
        solution = true;
      }
    }
  }
  cout << (solution ? "YES" : "NO") << endl;
  return 0;
}
