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 l, n, m, u, v;
int in[510], in1[510];
bool flag, ff;
struct node {
  int v, next;
} edge[100005];
int head[510];
void add(int u, int v) {
  l++;
  edge[l].v = v;
  edge[l].next = head[u];
  head[u] = l;
}
int main() {
  scanf("%d%d", &n, &m);
  memset(head, 0, sizeof(head));
  l = 0;
  memset(in, 0, sizeof(in));
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &u, &v);
    add(u, v);
    in[v]++;
  }
  for (int i = 1; i <= n; i++) {
    flag = 0;
    if (in[i]) {
      flag = 1;
      in[i]--;
    }
    memcpy(in1, in, sizeof(in));
    ff = 1;
    for (int j = 1; j <= n; j++) {
      int k = 1;
      while (k <= n && in1[k]) k++;
      if (k == n + 1) {
        ff = 0;
        break;
      }
      in1[k] = 0x3f3f3f3f;
      for (int p = head[k]; p; p = edge[p].next) in1[edge[p].v]--;
    }
    if (flag) in[i]++;
    if (ff) {
      printf("YES");
      break;
    }
  }
  if (!ff) printf("NO");
}
