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;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
inline long long toll(string s) {
  long long v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
int n, m;
vector<int> adj[100000 + 5];
vector<pair<int, int>> test;
int vis[100000 + 5];
bool done;
void dfs(int u) {
  vis[u] = 1;
  if (done) return;
  for (int v : adj[u]) {
    if (vis[v] == 2) continue;
    if (vis[v] == 1) done = true;
    test.push_back({u, v});
    if (!vis[v]) dfs(v);
    if (done) return;
    test.pop_back();
  }
  vis[u] = 2;
}
int nf, ns;
bool can = 0;
void check(int u) {
  vis[u] = 1;
  for (int v : adj[u]) {
    if (vis[v] == 2 || (u == nf && v == ns)) continue;
    if (vis[v] == 1) can = 1;
    if (vis[v] == 0) check(v);
  }
  vis[u] = 2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v);
  }
  for (int i = 0; i < n; i++)
    if (!vis[i]) dfs(i);
  if (!done)
    cout << "YES\n";
  else {
    for (pair<int, int> p : test) {
      nf = p.first;
      ns = p.second;
      memset(vis, 0, sizeof(vis));
      can = 0;
      for (int i = 0; i < n; i++)
        if (!vis[i]) check(i);
      if (!can) {
        cout << "YES\n";
        return 0;
      }
    }
    cout << "NO\n";
  }
  return 0;
}
