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;
long long n;
long long m;
vector<pair<long long, long long> > g[505];
long long vis[505];
long long parentedge[505];
long long U[500005], V[500005];
vector<long long> store;
vector<pair<long long, long long> > edges;
bool findloop(long long u, long long edgeidx) {
  if (vis[u] == 1) {
    store.push_back(edgeidx);
    long long src = U[edgeidx];
    while (src != u) {
      long long eidx = parentedge[src];
      store.push_back(eidx);
      src = U[eidx];
    }
    return 1;
  }
  if (vis[u] == 2) return 0;
  vis[u] = 1;
  parentedge[u] = edgeidx;
  for (auto v : g[u]) {
    if (findloop(v.first, v.second)) return 1;
  }
  vis[u] = 2;
  return 0;
}
bool checkloop() {
  for (long long i = 1; i <= n; i++) {
    if (!vis[i] && findloop(i, -1)) return 1;
  }
  return 0;
}
bool findloop2(long long u, long long removed) {
  if (vis[u] == 1) return 1;
  if (vis[u] == 2) return 0;
  vis[u] = 1;
  for (auto it : g[u]) {
    if (it.second == removed) continue;
    if (findloop2(it.first, removed)) return 1;
  }
  vis[u] = 2;
  return 0;
}
bool checkloop2(long long removed) {
  fill(vis + 1, vis + n + 1, 0);
  for (long long i = 1; i <= n; i++) {
    if (!vis[i] && findloop2(i, removed)) return 0;
  }
  return 1;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  cin >> m;
  for (long long i = 1; i <= m; i++) {
    long long x, y;
    cin >> x >> y;
    U[i] = x;
    V[i] = y;
    g[x].push_back({y, i});
  }
  if (!checkloop())
    cout << "YES\n";
  else {
    for (auto it : store) {
      if (checkloop2(it)) {
        cout << "YES\n";
        return 0;
      }
    }
    cout << "NO\n";
  }
  return 0;
}
