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 a[300005][2];
int b[300005][2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, x, y;
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> a[i][0] >> a[i][1];
  }
  for (int i = 0; i < 2; ++i) {
    int uk = 1;
    for (int j = 1; j <= m; ++j)
      if (a[j][0] != a[1][i] && a[j][1] != a[1][i])
        b[uk][0] = a[j][0], b[uk][1] = a[j][1], uk++;
    if (uk == 1) {
      if (a[1][i] > 1) {
        cout << "YES" << '\n';
        return 0;
      }
      if (a[1][i] < n) {
        cout << "YES" << '\n';
        return 0;
      }
      continue;
    }
    for (int p = 0; p < 2; ++p) {
      bool f = true;
      for (int j = 1; j < uk; ++j) {
        if (b[j][0] != b[1][p] && b[j][1] != b[1][p]) {
          f = false;
          break;
        }
      }
      if (f) {
        cout << "YES" << '\n';
        return 0;
      }
    }
  }
  cout << "NO" << '\n';
  return 0;
}
