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;
using ll = long long;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)1e5 + 5;
int n, m, p[505];
vector<int> e[505];
bool mark_temp[505], mark_perm[505];
bool found_cycl = 0;
int x;
void dfs(int pp, int u) {
  p[u] = pp;
  if (mark_temp[u]) {
    found_cycl = 1;
    x = u;
    return;
  }
  mark_temp[u] = 1;
  for (int i = 0; i < e[u].size() && !found_cycl; ++i)
    if (!mark_perm[e[u][i]]) dfs(u, e[u][i]);
  mark_temp[u] = 0;
  mark_perm[u] = 1;
}
void rm(pair<int, int>& edge) {
  e[edge.first].erase(
      find(e[edge.first].begin(), e[edge.first].end(), edge.second));
}
void rt(pair<int, int>& edge) { e[edge.first].push_back(edge.second); }
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    e[u].push_back(v);
  }
  for (int i = 1; i <= n; ++i) {
    if (!mark_perm[i] && !found_cycl) dfs(0, i);
  }
  if (!found_cycl)
    cout << "YES";
  else {
    int cycnode = x;
    vector<pair<int, int> > cedges;
    do {
      cedges.push_back({p[cycnode], cycnode});
      cycnode = p[cycnode];
    } while (cycnode != x);
    for (auto& edge : cedges) {
      rm(edge);
      found_cycl = 0;
      memset(mark_perm, 0, sizeof(mark_perm) * sizeof(bool));
      for (int i = 1; i <= n; ++i) {
        if (!mark_perm[i] && !found_cycl) dfs(0, i);
      }
      if (!found_cycl) {
        cout << "YES";
        break;
      }
      rt(edge);
    }
    if (found_cycl) cout << "NO";
  }
  return 0;
}
