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, d[510], a[510];
vector<int> g[510];
bool topsort(int x) {
  memcpy(a, d, sizeof(d));
  int num = 0;
  a[x]--;
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (a[i] == 0) q.push(i), num++;
  while (!q.empty()) {
    int i = q.front();
    q.pop();
    for (int j = 0; j < g[i].size(); j++)
      if (!(--a[g[i][j]])) q.push(g[i][j]), num++;
  }
  return num == n;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    d[v]++;
  }
  for (int i = 1; i <= n; i++)
    if (topsort(i)) {
      puts("YES");
      return 0;
    }
  puts("NO");
  return 0;
}
