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;
int main() {
  int t = 1;
  while (t--) {
    long long int i, y, h, j, k, r = 0, l = 0, n, m, x = 0, o = 0,
                                 cnt1 = INT_MAX, cnt2 = 0, cnt3 = 0, cnt4 = 0;
    cin >> n >> m;
    pair<long long int, long long int> a[m];
    for (i = 0; i < m; i++) {
      cin >> a[i].first >> a[i].second;
    }
    long long int b[2];
    b[0] = a[0].first;
    b[1] = a[0].second;
    for (i = 0; i < 2; i++) {
      o = 0;
      vector<long long int> c(n + 1, 0);
      for (j = 0; j < m; j++) {
        if (a[j].first != b[i] && a[j].second != b[i]) {
          c[a[j].first]++;
          c[a[j].second]++;
          o++;
        }
      }
      if (*max_element(c.begin(), c.end()) == o) {
        cout << "YES\n";
        return 0;
      }
    }
    cout << "NO\n";
  }
}
