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 comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
vector<set<long long> > adj;
long long vis[505];
long long temp[505], parent[505];
long long height[505];
long long mx = 0, ind, ind2;
bool flag = true;
void dfs(long long ver, long long x, long long p) {
  height[ver] = x;
  temp[ver] = p;
  vis[ver] = 1;
  for (auto i : adj[ver]) {
    if (vis[i] == 0) {
      dfs(i, x + 1, ver);
    } else if (abs(height[ver] - height[i]) > mx && vis[i] == 1) {
      mx = abs(height[ver] - height[i]);
      long long z = ver;
      ind = ver;
      ind2 = i;
      while (z != 0) {
        parent[z] = temp[z];
        z = temp[z];
      }
    }
  }
  vis[ver] = 2;
}
void dfs2(long long ver) {
  if (!flag) return;
  vis[ver] = 1;
  for (auto i : adj[ver]) {
    if (vis[i] == 0)
      dfs2(i);
    else if (vis[i] == 1) {
      flag = false;
      return;
    }
  }
  vis[ver] = 2;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  adj.resize(n + 1);
  long long x, y;
  for (long long i = 0; i < m; i++) {
    cin >> x >> y;
    adj[x].insert(y);
  }
  for (long long i = 1; i <= n; i++) {
    if (vis[i] == 0) dfs(i, 0, 0);
  }
  long long z = ind;
  if (mx == 0) {
    cout << "YES";
    return 0;
  }
  parent[ind2] = ind;
  long long cnt = 0;
  while (cnt <= mx) {
    cnt++;
    adj[parent[z]].erase(z);
    flag = true;
    fill_n(vis, 505, 0);
    for (long long i = 1; i <= n; i++) {
      if (vis[i] == 0) {
        dfs2(i);
      }
    }
    if (flag) {
      cout << "YES";
      return 0;
    }
    adj[parent[z]].insert(z);
    z = parent[z];
  }
  cout << "NO";
  clock_t clk;
  clk = clock();
  clk = clock() - clk;
  cerr << fixed << setprecision(6) << "Time: " << ((double)clk) / CLOCKS_PER_SEC
       << "\n";
  return 0;
}
