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;
const int maxn = 1e5 + 5;
vector<int> v[maxn];
int n, m, in[maxn];
int idx[maxn], out[maxn];
bool check(int x) {
  for (int i = 1; i <= n; i++) in[i] = idx[i];
  in[x]--;
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (in[i] == 0) q.push(i);
  }
  int num = 0;
  while (!q.empty()) {
    int s = q.front();
    q.pop();
    num++;
    for (int i = 0; i < out[s]; i++) {
      int t = v[s][i];
      in[t]--;
      if (in[t] == 0) {
        q.push(t);
      }
    }
  }
  return num == n;
}
void input() {
  cin >> n >> m;
  for (int i = 1, a, b; i <= m; i++) {
    cin >> a >> b;
    v[a].push_back(b);
    idx[b]++;
  }
  for (int i = 1; i <= n; i++) {
    out[i] = v[i].size();
  }
}
void work() {
  for (int i = 1; i <= n; i++) {
    if (check(i)) {
      puts("YES");
      return;
    }
  }
  puts("NO");
  return;
}
int main() {
  input();
  work();
  return 0;
}
