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 = 505;
const int eps = 1e-14;
const int mod = 1e9 + 7;
const long long inf = 1e18;
int n, m;
int g[maxn][maxn];
int deg[maxn], in[maxn];
bool topo() {
  queue<int> q;
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (in[i] == 0) q.push(i), cnt++;
  }
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (int i = 1; i <= n; i++) {
      if (g[now][i] && in[i]) {
        in[i]--;
        if (!in[i]) q.push(i), cnt++;
      }
    }
  }
  return cnt == n;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    g[x][y] = 1;
    deg[y]++;
  }
  int flag = 0;
  for (int i = 1; i <= n; i++) {
    if (deg[i]) {
      memcpy(in, deg, sizeof deg);
      in[i]--;
      if (topo()) {
        flag = 1;
        break;
      }
    }
  }
  printf(flag ? "YES\n" : "NO\n");
  return 0;
}
