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 = 500 + 7, maxd = 250000 + 7;
int n, m, deg[maxn], t[maxn];
vector<int> vec[maxn];
queue<int> qu;
bool is_ok() {
  int cnt = 0;
  for (int i = 1; i <= n; ++i) {
    if (t[i] == 0) qu.push(i), cnt++;
  }
  while (!qu.empty()) {
    int id = qu.front();
    qu.pop();
    for (auto i : vec[id]) {
      if (t[i]) {
        t[i]--;
        if (t[i] == 0) qu.push(i), cnt++;
      }
    }
  }
  return (cnt == n);
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    vec[u].push_back(v);
    deg[v]++;
  }
  for (int i = 1; i <= n; ++i) {
    if (deg[i]) {
      memcpy(t, deg, sizeof deg);
      t[i]--;
      if (is_ok()) {
        cout << "YES";
        return 0;
      }
    }
  }
  cout << "NO";
  return 0;
}
