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 = 505;
vector<vector<array<int, 2>>> v(N);
vector<int> vis(N), dis(N), low(N), block(N);
int cycle, edge;
vector<int> q, tmp, tmp2;
int tar = -1;
int dfs(int node, int p, int id) {
  vis[node] = 1;
  int f = 0;
  for (auto i : v[node]) {
    if (block[i[1]]) continue;
    if (vis[i[0]] == 1) {
      tar = i[0];
      tmp.push_back(i[1]);
      return 1;
    } else if (vis[i[0]] == 0) {
      int f1 = dfs(i[0], node, i[1]);
      if (f1) f = 1;
      if (f) {
        if (tar != -1) {
          tmp.push_back(i[1]);
          if (node == tar) tar = -1;
        }
        break;
      }
    }
  }
  vis[node] = 2;
  return f;
}
int detectCycle(int n) {
  vis = vector<int>(n + 1);
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    tar = -1;
    tmp.clear();
    int chk = dfs(i, -1, -1);
    if (chk) return 1;
  }
  return 0;
}
signed main() {
  int n, m, x, y;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    v[x].push_back({y, i});
  }
  cycle = detectCycle(n);
  if (cycle) {
    tmp2 = tmp;
    for (auto i : tmp2) {
      if (i == -1) continue;
      block[i] = 1;
      if (!detectCycle(n)) {
        cout << "YES";
        return 0;
      }
      block[i] = 0;
    }
  } else {
    cout << "YES";
    return 0;
  }
  cout << "NO";
}
