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 N = 1012;
int T, n, m, deg[N], tmp[N];
std::vector<int> g[N];
bool topsort() {
  static int q[N], l, r;
  l = 1, r = 0;
  for (int i = 1; i <= n; i++) {
    if (!deg[i]) {
      q[++r] = i;
    }
  }
  while (l <= r) {
    int u = q[l++];
    for (auto v : g[u]) {
      --deg[v];
      if (!deg[v]) {
        q[++r] = v;
      }
    }
  }
  return r == n;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    std::cin >> x >> y;
    tmp[y]++;
    g[x].push_back(y);
  }
  bool can = false;
  for (int i = 1; i <= n; i++) {
    memcpy(deg, tmp, n + 1 << 2);
    --deg[i];
    if (topsort()) {
      can = true;
      break;
    }
  }
  std::cout << (can ? "YES" : "NO") << "\n";
  return 0;
}
