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 = 555555;
int n, m, beg, endc;
vector<int> g[N], cycle;
int pr[N], us[N];
inline bool dfs(int v) {
  us[v] = 1;
  for (auto to : g[v]) {
    if (us[to] == 0) {
      pr[to] = v;
      if (dfs(to)) {
        return 1;
      }
    }
    if (us[to] == 1) {
      beg = to;
      endc = v;
      return 1;
    }
  }
  us[v] = 2;
  return false;
}
int main() {
  ;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1, x, y; i <= m; ++i) {
    cin >> x >> y;
    g[x].push_back(y);
  }
  bool t = 0;
  for (int i = 1; i <= n; ++i) {
    fill(us + 1, us + 1 + n, 0);
    if (dfs(i)) {
      t = 1;
      break;
    }
  }
  if (!t) {
    return puts("YES"), NULL;
  }
  for (int i = endc; i != beg; i = pr[i]) {
    cycle.push_back(i);
  }
  cycle.push_back(beg);
  reverse(cycle.begin(), cycle.end());
  for (int i = 0; i < cycle.size(); ++i) {
    int v = cycle[i];
    for (int j = 0; j < g[v].size(); ++j) {
      fill(us + 1, us + 1 + n, 0);
      sort(g[v].begin(), g[v].end());
      int to = g[v][j];
      bool t = 0;
      g[v].erase(g[v].begin() + j);
      for (int k = 1; k <= n; ++k) {
        if (dfs(k)) {
          t = 1;
          break;
        }
      }
      if (!t) {
        return puts("YES"), NULL;
      }
      g[v].push_back(to);
    }
  }
  puts("NO");
  return 0;
}
