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 = (long long)1e18;
const long long mod = (long long)1e9 + 7;
const double eps = (double)1e-9;
const double pi = acos(-1.0);
const int dx[] = {0, 0, 1, 0, -1};
const int dy[] = {0, 1, 0, -1, 0};
const int N = 505;
bool found;
bool bad[N][N];
vector<pair<int, int> > e;
vector<int> g[N];
int n, m, u[N], pr[N];
bool get(int v) {
  u[v] = 1;
  for (auto to : g[v]) {
    if (!u[to]) {
      pr[to] = v;
      if (get(to)) return 1;
    } else if (u[to] == 1) {
      int u = v;
      e.push_back({v, to});
      while (u != to) {
        e.push_back({pr[u], u});
        u = pr[u];
      }
      return 1;
    }
  }
  u[v] = 2;
  return 0;
}
bool check(int v) {
  u[v] = 1;
  for (auto to : g[v]) {
    if (bad[v][to]) continue;
    if (!u[to] && !check(to))
      return 0;
    else if (u[to] == 1)
      return 0;
  }
  u[v] = 2;
  return 1;
}
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
  }
  for (int i = 1; i <= n; i++) {
    if (!u[i] && get(i)) {
      found = 1;
      break;
    }
  }
  if (!found) {
    cout << "YES";
    return 0;
  }
  for (auto i : e) {
    bad[i.first][i.second] = 1;
    memset(u, 0, sizeof(u));
    bool ok = 1;
    for (int j = 1; j <= n; j++)
      if (!u[j]) ok &= (check(j));
    if (ok) {
      cout << "YES";
      return 0;
    }
    bad[i.first][i.second] = 0;
  }
  cout << "NO";
  return 0;
}
