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;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, m;
  cin >> n >> m;
  long long int a[4];
  vector<pair<long long int, long long int> > v;
  for (long long int i = 0; i < m; i++) {
    long long int c, b;
    cin >> c >> b;
    v.push_back(make_pair(c, b));
  }
  a[0] = v[0].first, a[1] = v[0].second;
  long long int flag = 0;
  for (long long int i = 1; i < m; i++) {
    if (v[i].first != a[0] && v[i].second != a[1] && v[i].first != a[1] &&
        v[i].second != a[0]) {
      a[2] = v[i].first;
      a[3] = v[i].second;
      flag = 1;
      break;
    }
  }
  if (!flag) {
    cout << "YES";
    return 0;
  } else {
    for (long long int i = 0; i < 4; i++) {
      for (long long int j = i + 1; j < 4; j++) {
        long long int count = 0;
        for (long long int k = 0; k < m; k++)
          if (v[k].first == a[i] || v[k].second == a[i] || v[k].first == a[j] ||
              v[k].second == a[j])
            count++;
        if (count == m) {
          cout << "YES";
          return 0;
        }
      }
    }
  }
  cout << "NO";
  return 0;
}
