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 stu[501], nxt[200010], vto[200010], tot = 0;
void added(int u, int v) { vto[tot] = v, nxt[tot] = stu[u], stu[u] = tot++; }
int d[501], ud[501], vis[501];
int main(int argc, char const *argv[]) {
  int n, m, a, b;
  scanf("%d%d", &n, &m);
  memset(stu, -1, sizeof stu);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    added(a, b);
    d[b]++;
  }
  for (int i = 1; i <= n; i++) {
    if (d[i] == 0) continue;
    memset(vis, 0, sizeof vis);
    int t = 0;
    queue<int> q;
    for (int j = 1; j <= n; j++) {
      if (i == j)
        ud[j] = d[j] - 1;
      else
        ud[j] = d[j];
      if (ud[j] == 0) {
        vis[j] = 1;
        q.push(j);
        t++;
      }
    }
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (int p = stu[u]; ~p; p = nxt[p]) {
        int v = vto[p];
        ud[v]--;
        if (ud[v] == 0) q.push(v), t++;
      }
    }
    if (t == n) {
      printf("YES\n");
      return 0;
    }
  }
  queue<int> q;
  memset(vis, 0, sizeof vis);
  int t = 0;
  for (int j = 1; j <= n; j++) {
    if (d[j] == 0) {
      vis[j] = 1;
      q.push(j);
      t++;
    }
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int p = stu[u]; ~p; p = nxt[p]) {
      int v = vto[p];
      d[v]--;
      if (d[v] == 0) q.push(v), t++;
    }
  }
  if (t == n) {
    printf("YES\n");
    return 0;
  }
  printf("NO\n");
  return 0;
}
