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 MX = 501;
int n, m, m_u, m_v, c, par[MX], build_tree;
char state[MX];
vector<int> adj[MX];
int dfs(int u) {
  if (state[u] == 'b') return 0;
  c++;
  state[u] = 'g';
  for (int v : adj[u]) {
    if (u == m_u && v == m_v) continue;
    if (state[v] == 'b') continue;
    if (build_tree) par[v] = u;
    if (state[v] == 'g') return v;
    int is_cycle = dfs(v);
    if (is_cycle) return is_cycle;
  }
  state[u] = 'b';
  return 0;
}
int detect() {
  for (int i = 1; i <= n; i++) state[i] = 'w';
  for (int i = 1; i <= n; i++) {
    int is_cycle = dfs(i);
    if (is_cycle) return is_cycle;
  }
  return 0;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
  }
  build_tree = 1;
  int k = detect();
  build_tree = 0;
  int cur = k;
  do {
    cur = par[cur];
    m_u = par[cur];
    m_v = cur;
    if (!detect()) {
      cout << "YES";
      return 0;
    }
  } while (cur != k);
  cout << "NO";
  return 0;
}
