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;
const int maxn = 511;
vector<int> w[maxn];
int d1[maxn], d2[maxn];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    w[u].push_back(v);
    ++d1[v];
  }
  bool fg = false;
  for (int i = 1; i <= n; ++i) {
    if (d1[i] == 0) continue;
    for (int j = 1; j <= n; ++j) d2[j] = d1[j];
    --d2[i];
    queue<int> que;
    int ct = 0;
    for (int j = 1; j <= n; ++j) {
      if (!d2[j]) {
        que.push(j);
        ++ct;
      }
    }
    while (!que.empty()) {
      int u = que.front();
      que.pop();
      for (int v : w[u]) {
        if (--d2[v] == 0) {
          que.push(v);
          ++ct;
        }
      }
    }
    if (ct == n) {
      fg = true;
      break;
    }
  }
  puts(fg ? "YES" : "NO");
  return 0;
}
