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;
vector<vector<long long> > v;
void update_cyc(vector<long long>& cyc, vector<long long>& par, long long dest,
                long long sor) {
  while (dest != sor) {
    cyc.push_back(dest);
    dest = par[dest];
  }
  cyc.push_back(sor);
  reverse(cyc.begin(), cyc.end());
  return;
}
bool dfs(long long s, vector<long long>& vis, vector<long long>& par,
         vector<long long>& cyc, long long p) {
  vis[s] = 1;
  par[s] = p;
  bool val = false;
  for (long long i = 0; i < v[s].size(); i++) {
    long long u = v[s][i];
    if (!vis[u]) {
      val |= dfs(u, vis, par, cyc, s);
      if (val) return true;
    } else if (vis[u] == 1) {
      update_cyc(cyc, par, s, u);
      return true;
    }
  }
  vis[s] = 2;
  return val;
}
bool dfs1(long long s, vector<long long>& vis, long long a, long long b) {
  vis[s] = 1;
  bool val = false;
  for (long long i = 0; i < v[s].size(); i++) {
    long long u = v[s][i];
    if (s == a && b == u) continue;
    if (!vis[u]) {
      val |= dfs1(u, vis, a, b);
      if (val) return true;
    } else if (vis[u] == 1) {
      return true;
    }
  }
  vis[s] = 2;
  return val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m;
  cin >> n >> m;
  v.resize(n + 1);
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    v[a].push_back(b);
  }
  vector<long long> cyc;
  vector<long long> vis(n + 1), par(n + 1);
  bool cnd = false;
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) {
      cnd |= dfs(i, vis, par, cyc, -1);
      if (cnd) break;
    }
  }
  if (!cnd) {
    cout << "YES" << endl;
    return 0;
  }
  long long k = cyc.size();
  for (long long i = 0; i < k; i++) {
    vis.clear();
    vis.resize(n + 1);
    bool val = false;
    for (long long j = 1; j <= n; j++) {
      if (!vis[j]) {
        val |= dfs1(j, vis, cyc[i], cyc[(i + 1) % k]);
      }
    }
    if (!val) {
      cout << "YES" << endl;
      return 0;
    }
  }
  cout << "NO" << endl;
  return 0;
}
