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 = 1e5 + 10;
int n, m;
vector<int> G[maxn], ts;
int pos[maxn], x;
bool mark[maxn], fr;
void dfs(int a) {
  mark[a] = 1;
  for (auto b : G[a]) {
    if ((!mark[b]) && (fr || b != x)) {
      dfs(b);
    }
    if (b == x) fr = 1;
  }
  pos[a] = ts.size();
  ts.push_back(a);
}
int s[maxn], t[maxn];
int32_t main() {
  scanf("%d%d", &n, &m);
  for (int i = int(0); i < int(m); i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    s[i] = a;
    t[i] = b;
    G[a].push_back(b);
  }
  for (; x < n; x++) {
    fill(mark, mark + n, 0);
    fr = 0;
    for (int i = int(0); i < int(n); i++) {
      if (!mark[i]) dfs(i);
    }
    int ans = 0;
    for (int i = int(0); i < int(m); i++) {
      ans += (pos[s[i]] < pos[t[i]]);
    }
    if (ans < 2) {
      printf("YES\n");
      return 0;
    }
  }
  printf("NO\n");
}
