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 long long inf = 1e18;
const int N = 1e6 + 10;
const long long mod = 1e9 + 7;
const int base = 131;
const double pi = acos(-1);
const double eps = 1e-8;
inline long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return ans;
}
unordered_map<long long, long long> mp;
long long vis[N], num[N], did[N];
long long n, m, k, x, y, z;
long long a[N], b[N], in[N];
long long cx, cy, cnt, ans, sum, flag, t, ff, q, w;
vector<int> v[N], res;
set<string> st;
bool topo() {
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (!in[i]) q.push(i);
  }
  cnt = 0;
  while (q.size()) {
    int x = q.front();
    q.pop();
    cnt++;
    for (int to : v[x]) {
      in[to]--;
      if (in[to] == 0) q.push(to);
    }
  }
  return cnt == n;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    b[y]++;
  }
  for (int i = 1; i <= n; i++) {
    if (b[i] > 0) {
      b[i]--;
      for (int j = 1; j <= n; j++) {
        in[j] = b[j];
      }
      if (topo()) {
        return cout << "YES" << endl, 0;
      }
      b[i]++;
    }
  }
  cout << "NO";
}
