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>
const int maxm = 100001;
const int maxn = 505;
int g[maxm];
int p[maxm];
int l[maxn], k;
int n, m, u, v;
int c[maxn];
int t[maxn];
int b[maxn], lb;
void insert(int u, int v) {
  k++;
  g[k] = v;
  p[k] = l[u];
  l[u] = k;
}
void strcpy(int a[], int b[]) {
  for (int i = 1; i <= n; i++) a[i] = b[i];
}
int bfs() {
  int u, v, res = 0;
  for (int i = 0; i < lb; i++) {
    u = b[i];
    for (int i = l[u]; i; i = p[i]) {
      v = g[i];
      if (c[v] == 0) continue;
      c[v]--;
      if (c[v] == 0) b[lb++] = v, res++;
    }
  }
  return res;
}
int main() {
  int col = 0;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    insert(u, v);
    c[v]++;
  }
  for (int i = 1; i <= n; i++) {
    if (c[i] == 0) {
      col++;
      b[lb++] = i;
    }
  }
  col += bfs();
  if (col == n) return puts("YES"), 0;
  strcpy(t, c);
  for (int i = 1; i <= n; i++) {
    if (c[i] == 1) {
      c[i]--;
      lb = 0;
      b[lb++] = i;
      if (col + 1 + bfs() == n) return puts("YES"), 0;
      strcpy(c, t);
    }
  }
  puts("NO");
  return 0;
}
