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 N = 510;
int vis[N], p[N];
vector<int> g[N];
int n, m;
vector<pair<int, int> > e;
void dfs(int u) {
  vis[u] = 1;
  for (int v : g[u]) {
    if (!vis[v]) {
      p[v] = u;
      dfs(v);
    } else if (vis[v] == 1 and e.empty()) {
      for (int a = u; a != v; a = p[a]) e.emplace_back(p[a], a);
      e.emplace_back(u, v);
    }
  }
  vis[u] = 2;
}
int dfs2(int u, const pair<int, int>& ed) {
  vis[u] = 1;
  int ok = 1;
  for (int v : g[u]) {
    if (u == ed.first and v == ed.second) continue;
    if (!vis[v])
      ok = ok and dfs2(v, ed);
    else if (vis[v] == 1)
      ok = 0;
  }
  vis[u] = 2;
  return ok;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i);
      if (!e.empty()) break;
    }
  }
  if (e.empty()) {
    printf("YES\n");
    return 0;
  }
  for (pair<int, int> edge : e) {
    int ok = 1;
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; i++) {
      if (!vis[i]) ok = ok and dfs2(i, edge);
    }
    if (ok) {
      printf("YES\n");
      return 0;
    }
  }
  printf("NO\n");
  return 0;
}
