CodeForces 915D Almost Acyclic Graph

Description

You are given a 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 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 \(m\)

\(\left(2 \le n \le 500, 1 \le m \le \min\left(n \cdot\left(n - 1\right), 100000\right)\right)\) — 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 \(\left(1 \le u, v \le n, u \neq v\right)\). Each ordered pair \(\left(u, v\right)\) 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

Output

YES

Input

Output

NO

Note

In the first example you can remove edge \(2 \rightarrow 3\) and the graph becomes acyclic.

In the second example you have to remove at least two edges (for example, \(2 \rightarrow 1\) and \(2 \rightarrow 3\)) in order to make the graph acyclic.

题解

有向图无环当且仅当存在拓扑序，而删掉边\(\left(u, v\right)\)的作用是使点\(v\)的入度减一，尽管边的数量是\(100000\)，但是对于同一个顶点，删掉不同入边的效果是等价的，所以我们只需要枚举每个顶点，将其入度减一，检查是否存在拓扑序即可。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 511;
vector<int> w[maxn];
int d1[maxn], d2[maxn];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    w[u].push_back(v);
    ++d1[v];
  }
  bool fg = false;
  for (int i = 1; i <= n; ++i) {
    if (d1[i] == 0) continue;
    for (int j = 1; j <= n; ++j)
      d2[j] = d1[j];
    --d2[i];
    queue<int> que;
    int ct = 0;
    for (int j = 1; j <= n; ++j) {
      if (!d2[j]) {
        que.push(j);
        ++ct;
      }
    }
    while (!que.empty()) {
      int u = que.front();
      que.pop();
      for (int v : w[u]) {
        if (--d2[v] == 0) {
          que.push(v);
          ++ct;
        }
      }
    }
    if (ct == n) {
      fg = true;
      break;
    }
  }
  puts(fg ? "YES" : "NO");
  return 0;
}