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 double pi = acos(-1.0);
int const lmt = 1e5 + 5;
long long int n, m;
vector<int> adj[502], lin, ck(lmt, true);
vector<pair<int, int> > e(lmt);
bool cyc = false, ok = false;
int vis[502];
bool dfs2(int p) {
  vis[p] = 1;
  for (auto j : adj[p]) {
    if (!ck[j]) continue;
    if (!vis[e[j].second]) {
      if (dfs2(e[j].second)) return true;
    } else if (vis[e[j].second] == 1) {
      return true;
    }
  }
  vis[p] = 2;
  return false;
}
bool has_cycle() {
  memset(vis, 0, sizeof vis);
  for (int i = 1; i <= n; i++) {
    if (!vis[i] && dfs2(i)) return true;
  }
  return false;
}
void dfs1(int p) {
  if (cyc || ok) return;
  vis[p] = 1;
  for (auto j : adj[p]) {
    lin.push_back(j);
    if (!vis[e[j].second]) {
      dfs1(e[j].second);
    } else if (!cyc && vis[e[j].second] == 1) {
      int nod = e[j].second;
      for (int i = lin.size() - 1; i >= 0; i--) {
        ck[lin[i]] = false;
        if (!has_cycle()) ok = true;
        ;
        ck[lin[i]] = true;
        if (e[lin[i]].first == nod) {
          break;
        }
      }
      cyc = true;
      return;
    }
    lin.pop_back();
  }
  vis[p] = 2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  int u, v;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    adj[u].push_back(i);
    e[i] = {u, v};
  }
  for (int i = 1; !cyc && i <= n; i++) {
    if (!vis[i]) {
      dfs1(i);
    }
  }
  if (!cyc) {
    cout << "YES\n";
    return 0;
  }
  cout << (ok ? "YES" : "NO") << "\n";
}
