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 = 501;
vector<int> g[maxn];
int was[maxn], st = -1, p[maxn];
vector<int> cycle;
int dfs(int v, int s, int f, int par) {
  p[v] = par;
  was[v] = 1;
  int ans = 0;
  for (int x : g[v]) {
    if (v != s or x != f) {
      if (was[x] == 1)
        ans++, p[x] = v, st = x;
      else if (was[x] == 0)
        ans += dfs(x, s, f, v);
    }
  }
  was[v] = 2;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  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++) {
    if (!was[i]) dfs(i, -1, -1, -1);
  }
  if (st == -1) {
    cout << "YES" << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) st = p[st];
  cycle.push_back(st);
  st = p[st];
  while (st != cycle[0]) cycle.push_back(st), st = p[st];
  cycle.push_back(cycle[0]);
  for (int i = 1; i < cycle.size(); i++) {
    fill(was, was + n + 1, 0);
    int ans = 0;
    for (int j = 1; j <= n; j++) {
      if (!was[j]) ans += dfs(j, cycle[i], cycle[i - 1], -1);
    }
    if (ans == 0) {
      cout << "YES" << endl;
      return 0;
    }
  }
  cout << "NO" << endl;
  return 0;
}
