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 MAXN = 600;
vector<pair<int, int>> st;
vector<pair<int, int>> eds[MAXN];
int was[MAXN];
int n, m;
vector<pair<int, int>> vv;
int gx;
int dfs1(int v) {
  was[v] = 1;
  for (auto e : eds[v]) {
    int u = e.first;
    if (!was[u]) {
      st.push_back(make_pair(u, e.second));
      if (dfs1(u)) {
        return 1;
      }
    } else if (was[u] == 1) {
      reverse(st.begin(), st.end());
      while (st.back().first != u) st.pop_back();
      st.back().second = e.second;
      reverse(st.begin(), st.end());
      vv = st;
      return 1;
    }
  }
  was[v] = 2;
  return 0;
}
int dfs2(int v) {
  was[v] = 1;
  for (auto e : eds[v]) {
    int u = e.first;
    if (gx == e.second || was[u] == 2) continue;
    if (was[u] == 1) return 1;
    if (dfs2(u)) return 1;
  }
  was[v] = 2;
  return 0;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    eds[a].push_back({b, i});
  }
  for (int i = 0; i < n; ++i) {
    if (was[i]) continue;
    st.clear();
    st.push_back(make_pair(i, 0));
    if (dfs1(i)) {
      break;
    }
  }
  if (!vv.size()) {
    cout << "YES\n";
    return 0;
  }
  for (int i = 0; i < vv.size(); ++i) {
    gx = vv[i].second;
    int fl = 0;
    memset(was, 0, sizeof(was[0]) * n);
    for (int j = 0; j < n; ++j) {
      if (!was[j]) {
        if (dfs2(j)) {
          fl = 1;
          break;
        }
      }
    }
    if (!fl) {
      cout << "YES\n";
      return 0;
    }
  }
  cout << "NO\n";
  return 0;
}
