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;
int get() {
  int x = 0;
  char s = getchar();
  while (s < '0' || s > '9') s = getchar();
  while (s >= '0' && s <= '9')
    x = (x << 3) + (x << 1) + (s ^ 48), s = getchar();
  return x;
}
const int N = 505, M = 1e5 + 5;
int n, m, head[N], cntedge, in[N], IN[N];
struct edge {
  int to, nxt;
  edge() {}
  edge(const int T, const int NX) { to = T, nxt = NX; }
} e[M];
void add(const int u, const int v) {
  e[++cntedge] = edge(v, head[u]), head[u] = cntedge;
}
bool bfs(const int del) {
  for (int i = 1; i <= n; ++i) in[i] = IN[i];
  --in[del];
  queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (!in[i]) q.push(i);
  int cnt = 0;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    ++cnt;
    for (int i = head[x]; i; i = e[i].nxt)
      if (!(--in[e[i].to])) q.push(e[i].to);
  }
  return cnt == n;
}
int main() {
  n = get(), m = get();
  for (int i = 1, u, v; i <= m; ++i) u = get(), v = get(), add(u, v), ++IN[v];
  bool flg = 0;
  for (int i = 1; i <= n; ++i)
    if (IN[i] && bfs(i)) {
      flg = 1;
      break;
    }
  printf(flg ? "YES" : "NO");
  return 0;
}
