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 N = 500;
int n, m;
vector<int> g[N];
int state[N];
vector<int> c;
void dfs(int v) {
  state[v] = 1;
  for (int i = 0; i < g[v].size(); ++i)
    if (state[g[v][i]] != 2) {
      if (state[g[v][i]] == 1) {
        c.push_back(g[v][i]);
        c.push_back(v);
        return;
      } else {
        dfs(g[v][i]);
        if (!c.empty()) {
          if (state[g[v][i]] == 2)
            state[v] = 2;
          else {
            if (v == c.front())
              state[v] = 2;
            else
              c.push_back(v);
          }
          return;
        }
      }
    }
  state[v] = 2;
}
bool dfs2(int v) {
  state[v] = 1;
  for (int i = 0; i < g[v].size(); ++i) {
    if (state[g[v][i]] == 1)
      return true;
    else if (!state[g[v][i]])
      if (dfs2(g[v][i])) return true;
  }
  state[v] = 2;
  return false;
}
bool ch() {
  memset(state, 0, 4 * n);
  for (int i = 0; i < n; ++i)
    if (!state[i] && dfs2(i)) return false;
  return true;
}
int main() {
  scanf("%d%d", &n, &m);
  int a, b;
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &a, &b);
    g[a - 1].push_back(b - 1);
  }
  for (int i = 0; i < n && c.empty(); ++i)
    if (!state[i]) dfs(i);
  if (c.empty())
    printf("YES\n");
  else {
    reverse(c.begin(), c.end());
    for (int i = 0; i < c.size(); ++i) {
      g[c[i]].erase(find(g[c[i]].begin(), g[c[i]].end(),
                         i == c.size() - 1 ? c[0] : c[i + 1]));
      if (ch()) {
        printf("YES\n");
        return 0;
      }
      g[c[i]].push_back(i == c.size() - 1 ? c[0] : c[i + 1]);
    }
    printf("NO\n");
  }
  return 0;
}
