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;
bitset<512> cur, was, is[512];
vector<int> G[512];
vector<pair<int, int> > edges;
bool check(int v) {
  was[v] = cur[v] = true;
  for (int u : G[v])
    if (is[v][u])
      if (cur[u])
        return true;
      else if (!was[u] && check(u))
        return true;
  return cur[v] = false;
}
int n;
bool solve() {
  was.reset();
  cur.reset();
  for (int i(1); i <= n; ++i)
    if (!was[i] && check(i)) return false;
  return true;
}
bool find_cycle(int v) {
  was[v] = cur[v] = true;
  for (int u : G[v])
    if (cur[u] || (!was[u] && find_cycle(u))) {
      edges.push_back(make_pair(v, u));
      return true;
    }
  return cur[v] = false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int m;
  cin >> n >> m;
  for (int u, v; m; --m) {
    cin >> u >> v;
    G[u].push_back(v);
    is[u][v] = true;
  }
  for (int i(1); i <= n; ++i)
    if (!was[i] && find_cycle(i)) break;
  if (!edges.empty()) {
    for (pair<int, int> e : edges) {
      is[e.first][e.second] = false;
      if (solve()) {
        cout << "YES";
        return 0;
      }
      is[e.first][e.second] = true;
    }
    cout << "NO";
  } else
    cout << "YES";
}
