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;
int n;
vector<int> e[505];
int deg[505], d[505], q[505], op, cl;
bool inq[505];
int main() {
  int m;
  scanf("%d%d", &n, &m);
  while (m--) {
    int x, y;
    scanf("%d%d", &x, &y);
    e[x].push_back(y);
    d[y]++;
  }
  for (int Q = 1; Q <= n; Q++) {
    memcpy(deg, d, sizeof(deg));
    memset(inq, 0, sizeof(inq));
    op = cl = 0;
    for (int i = 1; i <= n; i++)
      if (deg[i] == 0) {
        q[op++] = i;
        inq[i] = true;
      }
    bool bad = false;
    while (cl < n) {
      if (op == cl) {
        if (deg[Q] == 1 && !inq[Q]) {
          q[op++] = Q;
          inq[Q] = true;
        } else {
          bad = true;
          break;
        }
      }
      int x = q[cl++];
      for (vector<int>::iterator it = e[x].begin(); it != e[x].end(); it++) {
        int y = *it;
        --deg[y];
        if (deg[y] == 0 && !inq[y]) {
          inq[y] = true;
          q[op++] = y;
        }
      }
    }
    if (!bad) {
      puts("YES");
      return 0;
    }
  }
  puts("NO");
  return 0;
}
