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 N = 500 + 3;
int n, m, vis[N], pre[N], st;
vector<int> adj[N], V;
pair<int, int> ban;
bool cycle = false;
void dfs(int u) {
  if (cycle) return;
  vis[u] = 1;
  for (int v : adj[u])
    if (pair<int, int>(u, v) != ban) {
      if (vis[v] != 2) pre[v] = u;
      if (!vis[v])
        dfs(v);
      else if (vis[v] == 1) {
        cycle = true;
        st = v;
      }
      if (cycle) return;
    }
  vis[u] = 2;
}
void init() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= (int)m; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
  }
}
void solve() {
  for (int i = 1; i <= (int)n; ++i)
    if (!vis[i]) dfs(i);
  if (!cycle) {
    cout << "YES";
    return;
  }
  V.push_back(st);
  for (int v = pre[st]; v != st; v = pre[v]) V.push_back(v);
  V.push_back(st);
  reverse(V.begin(), V.end());
  for (int i = 1; i <= (int)V.size() - 1; ++i) {
    ban = {V[i - 1], V[i]};
    memset(vis, 0, sizeof vis);
    cycle = false;
    for (int i = 1; i <= (int)n; ++i)
      if (!vis[i]) dfs(i);
    if (!cycle) {
      cout << "YES";
      return;
    }
  }
  cout << "NO";
}
int main() {
  init();
  solve();
}
