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> nei[501];
int ideg[501];
bool outcyc[501];
int main() {
  int n, m, i;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    nei[x].push_back(y);
  }
  for (i = 1; i <= n; i++) {
    int j, k, outcycno = 0;
    fill(ideg + 1, ideg + n + 1, 0);
    for (j = 1; j <= n; j++)
      for (k = 0; k < nei[j].size(); k++) ideg[nei[j][k]]++;
    fill(outcyc + 1, outcyc + n + 1, false);
    while (true) {
      bool changed = false;
      for (j = 1; j <= n; j++)
        if (ideg[j] == 0 || j == i && ideg[j] == 1) {
          if (outcyc[j]) continue;
          outcycno++;
          outcyc[j] = changed = true;
          for (k = 0; k < nei[j].size(); k++) ideg[nei[j][k]]--;
        }
      if (!changed) break;
    }
    if (outcycno == n) return !printf("YES");
  }
  printf("NO");
  return 0;
}
