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;
int n, m, x, y;
vector<vector<int>> a;
vector<vector<int>> g;
vector<bool> used(505);
void dfs(int v, vector<int> s) {
  used[v] = true;
  for (int u : g[v]) {
    if (!used[u]) {
      s.push_back(u);
      dfs(u, s);
      s.pop_back();
    } else if (u == x) {
      s.push_back(u);
      a.push_back(s);
      s.pop_back();
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  g.resize(n + 1);
  for (int i = 0; i < m; ++i) {
    cin >> x >> y;
    g[x].push_back(y);
  }
  for (int i = 1; i <= n; ++i) {
    x = i;
    for (int j = 1; j <= n; ++j) used[j] = false;
    dfs(x, {x});
  }
  unordered_map<int, int> mm;
  if (a.size() == 0) {
    cout << "YES";
    return 0;
  }
  for (auto s : a) {
    for (int i = 0; i < s.size() - 1; ++i) {
      ++mm[(s[i] << 16) + s[i + 1]];
    }
  }
  for (auto i : mm) {
    if (i.second == a.size()) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
  return 0;
}
