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 long long N = 505;
long long n, m, pr[N], deg[N];
bool d[N][N], used[N];
vector<long long> a;
void dfs(long long u) {
  used[u] = true;
  for (long long i = 1; i <= n; ++i)
    if (!used[i] && d[u][i])
      if (--deg[i] <= 0) dfs(i);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  while (m--) {
    long long x, y;
    cin >> x >> y;
    d[x][y] = true;
    pr[y]++;
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) deg[j] = pr[j], used[j] = false;
    deg[i] -= 1;
    for (long long j = 1; j <= n; j++)
      if (!used[j] && deg[j] <= 0) dfs(j);
    bool check = false;
    for (long long j = 1; j <= n; j++)
      if (!used[j]) {
        check = true;
        break;
      }
    if (!check) {
      cout << "yes";
      return 0;
    }
  }
  cout << "no";
}
