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 n, m;
vector<vector<int> > g;
vector<pair<int, int> > mi;
bool vis[1000000], p[1000000], ok = false;
int cont = 0;
void dfs(int u) {
  vis[u] = true;
  p[u] = true;
  for (int i = 0; i < g[u].size(); ++i) {
    if (!vis[g[u][i]]) {
      mi.push_back({u, g[u][i]});
      dfs(g[u][i]);
    } else if (p[g[u][i]])
      cont++;
  }
  p[u] = false;
}
void dfs2(int x, int y, int u) {
  vis[u] = true;
  p[u] = true;
  for (int i = 0; i < g[u].size(); ++i) {
    if (u == x && y == g[u][i]) continue;
    if (!vis[g[u][i]])
      dfs2(x, y, g[u][i]);
    else if (p[g[u][i]])
      ok = true;
  }
  p[u] = false;
}
int main() {
  cin >> n >> m;
  int x, y;
  g.assign(n + 5, vector<int>());
  memset(vis, false, sizeof vis);
  memset(p, false, sizeof p);
  for (int i = 0; i < m; ++i) {
    cin >> x >> y;
    x--, y--;
    g[x].push_back(y);
  }
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      dfs(i);
    }
  }
  if (cont <= 1) {
    cout << "YES" << endl;
    return 0;
  }
  for (int i = 0; i < mi.size(); ++i) {
    memset(vis, false, sizeof vis);
    memset(p, false, sizeof p);
    ok = false;
    for (int j = 0; j < n; ++j) {
      if (vis[j] == false) {
        dfs2(mi[i].first, mi[i].second, j);
      }
    }
    if (!ok) {
      cout << "YES" << endl;
      return 0;
    }
  }
  if (!ok)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
