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 INF = 1e9;
const int maxn = 550 + 5;
const long long mod = 1e9 + 7;
vector<int> G[maxn];
int dep[maxn];
int now[maxn];
int n, m;
bool topo() {
  queue<int> Q;
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (now[i] <= 0) Q.push(i), cnt++;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (auto i : G[u]) {
      --now[i];
      if (now[i] == 0) Q.push(i), cnt++;
    }
  }
  return cnt == n;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    dep[v]++;
  }
  bool flag = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) now[j] = dep[j];
    now[i]--;
    flag |= topo();
  }
  if (flag)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
