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;
vector<pair<int, int>> pr;
int pairs(int x, int y) {
  for (pair<int, int> p : pr) {
    if (p.first != x && p.second != x && p.first != y && p.second != y)
      return 0;
  }
  return 1;
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    pr.push_back(make_pair(a, b));
  }
  for (int i : {pr[0].first, pr[0].second}) {
    if (pairs(i, 0) == 1) {
      cout << "YES";
      exit(0);
    }
    for (int j = 1; j < m; j++) {
      if (pr[j].first != i && pr[j].second != i) {
        if (pairs(i, pr[j].first) == 1 || pairs(i, pr[j].second) == 1) {
          cout << "YES";
          exit(0);
        }
        break;
      }
    }
  }
  cout << "NO";
  return 0;
}
