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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int n, m, x, p[500005];
int w[500005], u[500005], mom[500005], timer = 1;
vector<int> cyc;
set<int> g[500005];
vector<pair<int, int>> e;
void dfs(int v, int par) {
  u[v] = 1;
  p[v] = par;
  for (auto to : g[v]) {
    p[to] = v;
    if (u[to] == 0)
      dfs(to, v);
    else if (u[to] == 1 && !cyc.size()) {
      int cur = v;
      cyc.push_back(cur);
      while (cur != to && p[cur] != cur) {
        cur = p[cur];
        cyc.push_back(cur);
      }
      reverse(cyc.begin(), cyc.end());
      cyc.push_back(cyc.front());
    }
  }
  u[v] = 2;
}
void have(int v) {
  u[v] = 1;
  for (auto to : g[v]) {
    if (u[to] == 0)
      have(to);
    else if (u[to] == 1) {
      x = 1;
    }
  }
  u[v] = 2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cout.precision(30);
  cerr.precision(7);
  ;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int v1, v2;
    cin >> v1 >> v2;
    g[v1].insert(v2);
  }
  for (int i = 1; i <= n; i++)
    if (!u[i]) {
      dfs(i, 0), timer++;
    };
  if (!cyc.size()) {
    cout << "YES\n";
    return 0;
  }
  for (int i = 1; i < cyc.size(); i++) {
    e.push_back({cyc[i - 1], cyc[i]});
  }
  for (auto r : e) {
    g[r.first].erase(r.second);
    x = 0;
    for (int i = 1; i <= n; i++) u[i] = 0;
    for (int i = 1; i <= n; i++)
      if (!u[i] && !x) have(i);
    if (!x) {
      cout << "YES\n";
      return 0;
    }
    g[r.first].insert(r.second);
  }
  cout << "NO\n";
  return 0;
}
