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 i, n, m, ans, x, l, r, tot, a[1000005], b[1000005], last[1000005],
    to[1000005], Next[1000005], q[1000005], y;
inline void add(int x, int y) {
  Next[++tot] = last[x];
  last[x] = tot;
  to[tot] = y;
}
inline bool check(int x) {
  for (int i = 1; i <= n; i++) a[i] = b[i];
  a[x]--;
  int rr = r, ll = l;
  q[++r] = x;
  while (l < r) {
    int k = q[++l];
    for (int i = last[k]; i; i = Next[i]) {
      a[to[i]]--;
      if (!a[to[i]]) q[++r] = to[i];
    }
  }
  if (r == n) return true;
  r = rr, l = ll;
  return false;
}
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= m; i++) scanf("%d%d", &x, &y), b[y]++, add(x, y);
  for (i = 1; i <= n; i++)
    if (b[i] == 0) q[++r] = i;
  while (l < r) {
    int k = q[++l];
    for (int i = last[k]; i; i = Next[i]) {
      b[to[i]]--;
      if (!b[to[i]]) q[++r] = to[i];
    }
  }
  if (r == n) return puts("YES"), 0;
  for (i = 1; i <= n; i++) {
    if (b[i] == 1) {
      if (check(i)) return puts("YES"), 0;
    }
  }
  puts("NO");
}
