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 = 4e5;
int mar[N], st[N], tin[N];
int fnd, n, m, sz;
vector<int> cycle, ed[N];
void dfs(int x, int d1, int d2) {
  mar[x] = 1;
  sz++;
  st[sz] = x;
  tin[x] = sz;
  for (int i = 0; i < ed[x].size(); i++) {
    int to = ed[x][i];
    if (x == d1 && to == d2) continue;
    if (mar[to] == 2) continue;
    if (mar[to] == 1) {
      fnd = 1;
      if (cycle.size() == 0) {
        for (int j = tin[to]; j <= sz; j++) cycle.push_back(st[j]);
      }
    } else
      dfs(to, d1, d2);
  }
  mar[x] = 2;
  sz--;
}
bool chck(int d1, int d2) {
  for (int i = 1; i <= n; i++) mar[i] = 0;
  fnd = 0;
  for (int i = 1; i <= n; i++)
    if (mar[i] == 0) dfs(i, d1, d2);
  return fnd;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    ed[u].push_back(v);
  }
  chck(0, 0);
  if (cycle.size() == 0) {
    cout << "YES\n";
    return 0;
  } else {
    cycle.push_back(cycle[0]);
    for (int i = 0; i + 1 < cycle.size(); i++)
      if (!chck(cycle[i], cycle[i + 1])) {
        cout << "YES\n";
        return 0;
      }
  }
  cout << "NO\n";
  return 0;
}
