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 = 503;
vector<int> v[MAXN];
vector<int> dospr;
int odw[MAXN];
int uk[MAXN];
int n, kon;
int pop, licz = 0;
bool last;
bool DFS(int x) {
  odw[x] = licz;
  for (auto it : v[x]) {
    if (odw[it] == licz && !uk[it]) {
      kon = it;
      dospr.push_back(x);
      return true;
    } else if (!odw[it]) {
      if (DFS(it)) {
        if (!last) dospr.push_back(x);
        if (x == kon) last = true;
        return true;
      }
    }
  }
  uk[x] = licz;
  return false;
}
bool DFS2(int x) {
  odw[x] = licz;
  for (auto it : v[x]) {
    if (odw[it] == licz && uk[it] <= pop)
      return false;
    else if (odw[it] <= pop) {
      if (!DFS2(it)) return false;
    }
  }
  uk[x] = licz;
  return true;
}
bool sprawdz(int x, int y) {
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i] == y) {
      swap(v[x][i], v[x].back());
      break;
    }
  }
  v[x].pop_back();
  pop = licz;
  for (int i = 1; i <= n; i++) {
    if (odw[i] <= pop) {
      licz++;
      if (!DFS2(i)) {
        v[x].push_back(y);
        return false;
      }
    }
  }
  v[x].push_back(y);
  return true;
}
int main() {
  int m;
  int x, y;
  scanf("%d %d", &n, &m);
  while (m--) {
    scanf("%d %d", &x, &y);
    v[x].push_back(y);
  }
  for (int i = 1; i <= n; i++) {
    if (!odw[i]) {
      licz++;
      if (DFS(i)) break;
    }
  }
  int roz = dospr.size() - 1;
  if (roz == -1) {
    printf("YES");
    return 0;
  }
  for (int i = 1; i <= roz; i++) {
    if (sprawdz(dospr[i], dospr[i - 1])) {
      printf("YES");
      return 0;
    }
  }
  if (sprawdz(dospr[0], dospr[roz]))
    printf("YES");
  else
    printf("NO");
  return 0;
}
