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 double PI = acos(-1.0);
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const int MOD = 1e8;
const int N = 5e2 + 5;
const int dir[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};
int e[N][N];
int in[N], cnt[N];
int n, m;
bool flag;
void init() {
  memset(e, 0, sizeof(e));
  memset(cnt, 0, sizeof(cnt));
  flag = 0;
}
bool topp() {
  queue<int> que;
  int num = 0;
  for (int i = 1; i < n + 1; i++)
    if (!in[i]) que.push(i), num++;
  while (!que.empty()) {
    int front = que.front();
    que.pop();
    for (int i = 1; i < n + 1; i++) {
      if (e[front][i] && in[i]) {
        in[i]--;
        if (!in[i]) que.push(i), num++;
      }
    }
  }
  return num == n;
}
int main() {
  ios::sync_with_stdio(false);
  while (cin >> n >> m) {
    init();
    for (int i = 0; i < m; i++) {
      int u, v;
      cin >> u >> v;
      e[u][v] = 1;
      cnt[v]++;
    }
    for (int i = 1; i < n + 1; i++) {
      if (!cnt[i]) continue;
      memcpy(in, cnt, sizeof(cnt));
      in[i]--;
      if (topp()) {
        flag = 1;
        cout << "YES" << endl;
        break;
      }
    }
    if (!flag) cout << "NO" << endl;
  }
  return 0;
}
