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;
int n, m;
vector<int> adj[1000];
stack<int> s;
int pos[1000];
bool vis[1000];
void dfs(int i) {
  vis[i] = 1;
  for (auto j : adj[i])
    if (!vis[j]) dfs(j);
  s.push(i);
}
int main() {
  scanf("%d", &n);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d", &x);
    scanf("%d", &y);
    adj[x].push_back(y);
  }
  bool y = 0;
  for (int i = 1; i <= n; i++) {
    int in = 1;
    for (int j = 1; j <= n; j++) vis[j] = 0;
    dfs(i);
    for (int j = 1; j <= n; j++)
      if (!vis[j]) dfs(j);
    while (s.size()) {
      pos[s.top()] = in++;
      s.pop();
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++)
      for (auto j : adj[i])
        if (pos[j] < pos[i]) cnt++;
    if (cnt <= 1) y = 1;
  }
  if (y)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}
