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, kol, cnt, all;
vector<int> qqq;
int used2[505];
int pred[505];
bool used[505];
vector<int> v[505];
vector<int> vv[505];
int clr[505];
vector<int> qq;
vector<int> q;
bool dfs3(int x, int y) {
  used2[x] = 1;
  pred[x] = y;
  for (int i = 0; i < v[x].size(); ++i) {
    int to = v[x][i];
    if (used2[to] == 2) continue;
    if (used2[to] == 0) {
      if (dfs3(to, x)) return true;
      continue;
    }
    while (x != to) {
      qqq.push_back(x);
      x = pred[x];
    }
    qqq.push_back(to);
    return true;
  }
  used2[x] = 2;
  return false;
}
bool dfs4(int x, int y, int xx, int yy) {
  used2[x] = 1;
  for (int i = 0; i < v[x].size(); ++i) {
    int to = v[x][i];
    if (x == xx && to == yy) continue;
    if (used2[to] == 2) continue;
    if (used2[to] == 0) {
      if (dfs4(to, x, xx, yy)) return true;
      continue;
    }
    return true;
  }
  used2[x] = 2;
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    vv[y].push_back(x);
  }
  for (int i = 1; i <= n; ++i)
    if (dfs3(i, -1)) break;
  if (qqq.size() == 0) {
    cout << "YES";
    return 0;
  }
  reverse(qqq.begin(), qqq.end());
  for (int i = 0; i < qqq.size(); ++i) {
    int x = qqq[i];
    int y;
    if (i < qqq.size() - 1)
      y = qqq[i + 1];
    else
      y = qqq[0];
    memset(used2, 0, sizeof(used2));
    bool t = false;
    for (int i = 1; i <= n; ++i)
      if (dfs4(i, -1, x, y)) {
        t = true;
        break;
      }
    if (t == false) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
}
