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 = 505;
int deg[maxn], ans[maxn], m, n;
vector<int> edge[maxn];
queue<int> que;
int topo() {
  int v, res = 0, crl[maxn];
  memset(crl, 0, sizeof(crl));
  for (int j = 1; j <= n; j++) {
    if (deg[j] == 0) {
      que.push(j);
    }
  }
  while (!que.empty()) {
    v = que.front();
    que.pop();
    ans[res++] = v;
    for (int j = 0; j < edge[v].size(); j++) {
      deg[edge[v][j]]--;
      crl[edge[v][j]]++;
      if (deg[edge[v][j]] == 0) que.push(edge[v][j]);
    }
  }
  for (int i = 1; i <= n; i++) {
    deg[i] += crl[i];
  }
  return res;
}
int main() {
  int tmp1, tmp2;
  scanf("%d%d", &n, &m);
  memset(deg, 0, sizeof(deg));
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &tmp1, &tmp2);
    edge[tmp1].push_back(tmp2);
    deg[tmp2]++;
  }
  for (int i = 0; i <= n; i++) {
    deg[i]--;
    tmp1 = topo();
    if (tmp1 == n) {
      puts("YES");
      return 0;
    }
    deg[i]++;
  }
  puts("NO");
  return 0;
}
