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 maxn = 510;
int clr[maxn], p[maxn], n, m, x, y, f, s;
bool a[maxn][maxn];
vector<int> ccl;
bool dfs(int v) {
  clr[v] = 1;
  for (int i = 1; i <= n; i++)
    if (a[v][i])
      if (clr[i] == 0) {
        p[i] = v;
        if (dfs(i)) return 1;
      } else if (clr[i] == 1) {
        f = v;
        s = i;
        return 1;
      }
  clr[v] = 2;
  return 0;
}
int main() {
  cin >> n >> m;
  while (m--) {
    cin >> x >> y;
    a[x][y] = 1;
  }
  for (int i = 1; i <= n; i++)
    if (clr[i] == 0)
      if (dfs(i)) {
        while (f != p[s]) {
          ccl.insert(ccl.begin(), f);
          f = p[f];
        }
        ccl.push_back(s);
        break;
      }
  if (!ccl.size()) {
    cout << "YES";
    return 0;
  }
  for (int i = 0; i < ccl.size() - 1; i++) {
    x = ccl[i];
    y = ccl[i + 1];
    a[x][y] = 0;
    bool ans = 1;
    for (int i = 1; i <= n; i++) clr[i] = 0;
    for (int i = 1; i <= n; i++)
      if (clr[i] == 0)
        if (dfs(i)) {
          ans = 0;
          break;
        }
    if (ans) {
      cout << "YES";
      return 0;
    }
    a[x][y] = 1;
  }
  cout << "NO";
}
