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;
enum { MOD0 = 1000000000, MOD7 = 1000000007, MOD9 = 1000000009 };
template <typename T>
ostream &operator<<(ostream &cout, vector<T> &a) {
  for (size_t i = 0; i < a.size(); ++i) cout << a[i] << " ";
  return cout;
}
template <typename T>
ostream &operator<<(ostream &cout, vector<vector<T> > &a) {
  for (size_t i = 0; i < a.size(); ++i) cout << a[i] << endl;
  return cout;
}
vector<int> g[500];
int ru, rv;
int cyc;
int used[500];
int p[500];
int n, m;
bool seted = 0;
bool dfs(int x) {
  used[x] = 1;
  for (int i = 0; i < g[x].size(); ++i) {
    int v = g[x][i];
    if (x == ru && v == rv) continue;
    if (used[v] == 1) {
      p[v] = x;
      cyc = v;
      return 1;
    } else if (!used[v]) {
      p[v] = x;
      if (dfs(v)) {
        return 1;
      }
    }
  }
  used[x] = 2;
  return 0;
}
bool check_cycle() {
  memset(used, 0, sizeof(used));
  for (int i = 0; i < n; ++i) {
    if (!used[i]) {
      if (dfs(i)) {
        return false;
      }
    }
  }
  return true;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    g[u - 1].push_back(v - 1);
  }
  ru = -1;
  rv = -1;
  if (check_cycle()) {
    cout << "YES" << '\n';
    return 0;
  }
  vector<int> q;
  q.push_back(cyc);
  int cur = cyc;
  do {
    cur = p[cur];
    q.push_back(cur);
  } while (cur != cyc);
  for (int i = 0; i < q.size(); ++i) {
    cur = q[i];
    ru = p[cur];
    rv = cur;
    if (check_cycle()) {
      cout << "YES" << '\n';
      return 0;
    }
  }
  cout << "NO" << '\n';
  return 0;
}
