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;
vector<int> E[505];
int used[505], pre[505], N;
vector<int> cyc;
int nx, ny;
bool go(int x, int p = -1) {
  pre[x] = p;
  used[x] = 1;
  for (int y : E[x]) {
    if (nx == x && y == ny) continue;
    if (used[y] == 1) {
      while (x != y) {
        cyc.push_back(x);
        x = pre[x];
      }
      cyc.push_back(y);
      return true;
    } else if (!used[y]) {
      if (go(y, x)) return true;
    }
  }
  used[x] = 2;
  return false;
}
bool hascycle() {
  cyc.clear();
  memset(used, 0, sizeof(used));
  for (int i = 1; i <= N; ++i)
    if (!used[i]) {
      if (go(i)) return true;
    }
  return false;
}
int main() {
  int M;
  scanf("%d%d", &N, &M);
  for (int i = 0; i < M; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    E[u].push_back(v);
  }
  if (hascycle()) {
    auto C = cyc;
    for (int i = 0; i < C.size(); ++i) {
      ny = C[i];
      nx = C[(i + 1) % C.size()];
      if (!hascycle()) {
        printf("YES\n");
        return 0;
      }
    }
    printf("NO\n");
  } else {
    printf("YES\n");
  }
}
