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 MAXM = 100010;
int U[MAXM], V[MAXM];
int n, m;
vector<pair<int, int> > neigh[MAXM];
int tk[MAXM], par[MAXM];
vector<int> special;
bool findLoop(int v, int p) {
  if (tk[v] == 1) {
    special.push_back(p);
    int u = U[p];
    while (u != v) {
      special.push_back(par[u]);
      u = U[par[u]];
    }
    return true;
  }
  if (tk[v] == 2) return false;
  tk[v] = 1;
  par[v] = p;
  for (int i = 0; i < neigh[v].size(); i++) {
    pair<int, int> u = neigh[v][i];
    if (findLoop(u.first, u.second)) return true;
  }
  tk[v] = 2;
  return false;
}
bool ok(int v, int forb) {
  if (tk[v] == 1) return false;
  if (tk[v] == 2) return true;
  tk[v] = 1;
  for (int i = 0; i < neigh[v].size(); i++) {
    pair<int, int> u = neigh[v][i];
    if (u.second == forb) continue;
    if (!ok(u.first, forb)) return false;
  }
  tk[v] = 2;
  return true;
}
bool check(int forb) {
  fill(tk, tk + n + 1, 0);
  for (int i = 1; i <= n; i++)
    if (!tk[i])
      if (!ok(i, forb)) return false;
  return true;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    U[i] = u, V[i] = v;
    neigh[u].push_back(make_pair(v, i));
  }
  for (int i = 1; i <= n; i++)
    if (!tk[i])
      if (findLoop(i, -1)) break;
  if (special.empty())
    cout << "YES" << endl;
  else {
    for (int i = 0; i < special.size(); i++)
      if (check(special[i])) {
        cout << "YES" << endl;
        return 0;
      }
    cout << "NO" << endl;
  }
  return 0;
}
