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>
unsigned n, in_Deg[505] = {0}, K[505];
std::vector<unsigned> V[505];
bool ok() {
  unsigned time = n, f;
  std::queue<unsigned> Q;
  for (unsigned i = 0; i < n; i++)
    if (!(K[i] = in_Deg[i])) time--, Q.push(i);
  if (!time) return true;
  while (!Q.empty()) {
    f = Q.front(), Q.pop();
    for (unsigned i = V[f].size() - 1; i + 1; i--)
      if (K[V[f][i]] && !(--K[V[f][i]])) {
        if (!(--time)) return true;
        Q.push(V[f][i]);
      }
  }
  return false;
}
int main() {
  unsigned m, u, v;
  bool yes = false;
  scanf("%u%u", &n, &m);
  while (m--) scanf("%u%u", &u, &v), V[u - 1].push_back(--v), in_Deg[v]++;
  for (unsigned i = 0; i < n && !yes; i++)
    if (in_Deg[i]) in_Deg[i]--, yes = ok(), in_Deg[i]++;
  puts(yes ? "YES" : "NO");
  return 0;
}
