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 inf = 1e5 + 10;
int n, m;
int tot, fi[inf], nxt[inf], to[inf], in[inf];
void link(int x, int y) {
  to[++tot] = y;
  nxt[tot] = fi[x];
  fi[x] = tot;
}
int que[inf], l, r;
int tmp[inf];
bool topo() {
  l = 1;
  r = 0;
  for (int i = 1; i <= n; i++) tmp[i] = in[i];
  for (int i = 1; i <= n; i++)
    if (!tmp[i]) que[++r] = i;
  while (l <= r) {
    int u = que[l++];
    for (int i = fi[u]; i; i = nxt[i]) {
      tmp[to[i]]--;
      if (!tmp[to[i]]) que[++r] = to[i];
    }
  }
  return r == n;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    link(x, y);
    in[y]++;
  }
  for (int i = 1; i <= n; i++) {
    if (!in[i]) continue;
    in[i]--;
    if (topo()) {
      cout << "YES" << endl;
      return 0;
    }
    in[i]++;
  }
  cout << "NO" << endl;
  return 0;
}
