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;
typedef struct Vertex {
  vector<int> adj;
  int color;
} Vertex;
vector<Vertex> vertices;
int backcnt = 0;
void dfsVisit(int v_ind) {
  if (backcnt >= 2) return;
  vertices[v_ind].color = 1;
  for (int i = 0; i < vertices[v_ind].adj.size(); i++) {
    if (vertices[vertices[v_ind].adj[i]].color == 0)
      dfsVisit(vertices[v_ind].adj[i]);
    else if (vertices[vertices[v_ind].adj[i]].color == 1)
      backcnt++;
  }
  vertices[v_ind].color = 2;
}
int main() {
  int n, m, u, v;
  cin >> n >> m;
  vertices.resize(n);
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    vertices[u - 1].adj.push_back(v - 1);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      vertices[j].color = 0;
    }
    backcnt = 0;
    dfsVisit(i);
    for (int j = 0; j < n; j++) {
      if (vertices[j].color == 0) {
        dfsVisit(j);
      }
    }
    if (backcnt <= 1) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
}
