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<vector<int> > g;
vector<char> used;
int countCycles = 0;
void dfs(int v) {
  used[v] = 1;
  int sz = g[v].size();
  for (int i = 0; i < sz; i++) {
    int to = g[v][i];
    if (used[to] == 0)
      dfs(to);
    else if (used[to] == 1)
      countCycles++;
  }
  used[v] = 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  g.resize(n + 1);
  used.resize(n + 1);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
  }
  for (int i = 1; i <= n; i++) {
    used.assign(n + 1, 0);
    countCycles = 0;
    dfs(i);
    for (int j = 1; j <= n; j++) {
      if (used[j] == 0) dfs(j);
    }
    if (countCycles <= 1) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
  return 0;
}
