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 = 505;
int n, m, foundCycle = 0, esv, cec = 0, mode = 1;
set<int> g[N];
vector<pair<int, int>> cel;
stack<int> fcs;
char vis[N];
void fc(int u) {
  vis[u] = 1;
  if (mode) fcs.push(u);
  for (int v : g[u]) {
    if (vis[v] == 0)
      fc(v);
    else if (vis[v] == 1) {
      foundCycle = 1, esv = v;
      return;
    }
    if (foundCycle) return;
  }
  vis[u] = 2;
  if (mode) fcs.pop();
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    g[u].insert(v);
  }
  for (int i = 1; i <= n; i++)
    if (vis[i] == 0 && !foundCycle) fc(i);
  if (!foundCycle) {
    cout << "YES" << endl;
    return 0;
  }
  int tprev = esv;
  while (fcs.top() != esv) {
    int u = fcs.top();
    fcs.pop();
    cel.push_back({u, tprev});
    tprev = u;
  }
  cel.push_back({esv, tprev});
  mode = 0;
  for (auto e : cel) {
    g[e.first].erase(e.second);
    foundCycle = 0, memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; i++)
      if (vis[i] == 0 && !foundCycle) fc(i);
    if (!foundCycle) {
      cout << "YES" << endl;
      return 0;
    }
    g[e.first].insert(e.second);
  }
  cout << "NO" << endl;
  return 0;
}
