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 N = 500;
int n, m, mark[N], cnt = 0;
vector<int> g[N];
void dfs(int u);
int main() {
  scanf("%d%d", &n, &m);
  for (; m--;) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    g[u].emplace_back(v);
  }
  for (int u = 0; u < n; u++) {
    cnt = 0;
    fill(mark, mark + n, 0);
    dfs(u);
    for (int v = 0; v < n; v++) {
      if (!mark[v]) {
        dfs(v);
      }
    }
    if (cnt <= 1) {
      puts("YES");
      return 0;
    }
  }
  puts("NO");
  return 0;
}
void dfs(int u) {
  mark[u] = 1;
  for (auto v : g[u]) {
    if (!mark[v]) {
      dfs(v);
    } else if (mark[v] == 1) {
      cnt++;
    }
  }
  mark[u] = 2;
}
