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>
#pragma GCC optimize(3)
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 510, M = N * N * 2;
const long long mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template <typename T>
inline void read(T &a) {
  char c = getchar();
  T x = 0, f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  a = f * x;
}
int gcd(int a, int b) { return (b > 0) ? gcd(b, a % b) : a; }
int h[N], ne[M], e[M], idx;
int back[N];
int in[N];
int n, m;
bool topsort() {
  int res = 0;
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (!in[i]) q.push(i);
  while (q.size()) {
    int t = q.front();
    q.pop();
    res++;
    for (int i = h[t]; ~i; i = ne[i]) {
      int j = e[i];
      in[j]--;
      if (!in[j]) q.push(j);
    }
  }
  return res == n;
}
void add(int x, int y) { ne[idx] = h[x], e[idx] = y, h[x] = idx++; }
int main() {
  read(n), read(m);
  memset(h, -1, sizeof h);
  while (m--) {
    int x, y;
    read(x), read(y);
    add(x, y);
    in[y]++;
  }
  for (int i = 1; i <= n; i++) back[i] = in[i];
  bool flag = false;
  for (int i = 1; i <= n; i++) {
    in[i]--;
    flag |= topsort();
    for (int j = 1; j <= n; j++) in[j] = back[j];
  }
  if (flag)
    puts("YES");
  else
    puts("NO");
}
