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 = 500 + 10;
int ruDeg[maxn], inDeg[maxn];
vector<int> vec[maxn];
queue<int> q;
int n, m;
bool topsort() {
  while (!q.empty()) q.pop();
  for (int i = 1; i <= n; i++)
    if (inDeg[i] == 0) q.push(i);
  int num = 0;
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    ++num;
    for (int i = 0; i < vec[now].size(); i++) {
      if (--inDeg[vec[now][i]] == 0) q.push(vec[now][i]);
    }
  }
  if (num == n)
    return true;
  else
    return false;
}
int main() {
  cin >> n >> m;
  int u, v;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    vec[u].push_back(v);
    ruDeg[v]++;
    inDeg[v]++;
  }
  int flag = 0;
  for (int i = 1; i <= n; i++) {
    memcpy(inDeg, ruDeg, sizeof(inDeg));
    --inDeg[i];
    if (topsort()) {
      flag = 1;
      break;
    }
  }
  if (flag)
    puts("YES");
  else
    puts("NO");
  return 0;
}
