Toad Ivan has m pairs of integers, each integer is between 1 and n, inclusive. The pairs are (a_1, b_1), (a_2, b_2), …, (a_m, b_m). 

He asks you to check if there exist two integers x and y (1 ≤ x < y ≤ n) such that in each given pair at least one integer is equal to x or y.

Input

The first line contains two space-separated integers n and m (2 ≤ n ≤ 300 000, 1 ≤ m ≤ 300 000) — the upper bound on the values of integers in the pairs, and the number of given pairs.

The next m lines contain two integers each, the i-th of them contains two space-separated integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the integers in the i-th pair.

Output

Output "YES" if there exist two integers x and y (1 ≤ x < y ≤ n) such that in each given pair at least one integer is equal to x or y. Otherwise, print "NO". You can print each letter in any case (upper or lower).

Examples

Input


4 6
1 2
1 3
1 4
2 3
2 4
3 4


Output


NO


Input


5 4
1 2
2 3
3 4
4 5


Output


YES


Input


300000 5
1 2
1 2
1 2
1 2
1 2


Output


YES

Note

In the first example, you can't choose any x, y because for each such pair you can find a given pair where both numbers are different from chosen integers.

In the second example, you can choose x=2 and y=4.

In the third example, you can choose x=1 and y=2.
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  long long n, m;
  cin >> n >> m;
  map<long long, set<long long>> adj;
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].insert(b);
    adj[b].insert(a);
  }
  long long mx = INT_MIN, v = 0;
  map<long long, long long> deg;
  for (auto i : adj) {
    deg[i.first] = i.second.size();
    if (mx < i.second.size()) {
      mx = i.second.size();
      v = i.first;
    }
  }
  deg[v] = 0;
  for (auto i : adj[v]) {
    deg[i]--;
  }
  long long f = 0;
  for (long long i = 1; i <= n; i++) {
    if (deg[i] > 0) {
      f = 1;
      break;
    }
  }
  if (!f) {
    cout << "YES\n";
    return 0;
  }
  mx = INT_MIN, v = 0;
  for (long long i = 1; i <= n; i++) {
    if (mx < deg[i]) {
      mx = deg[i];
      v = i;
    }
  }
  deg[v] = 0;
  for (auto i : adj[v]) {
    deg[i]--;
  }
  f = 0;
  for (long long i = 1; i <= n; i++) {
    if (deg[i] > 0) {
      f = 1;
      break;
    }
  }
  if (!f) {
    cout << "YES\n";
    return 0;
  }
  cout << "NO\n";
  return 0;
}
