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;
const int maxn = 3e5 + 7;
int fir[maxn], sec[maxn];
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) scanf("%d %d", fir + i, sec + i);
  for (int flag = 1; flag < 3; flag++) {
    int ind = (flag == 1 ? fir[1] : sec[1]);
    int ind2, ok = 1, j;
    for (j = 2; j <= m; j++) {
      if (fir[j] == ind || sec[j] == ind) {
        continue;
      }
      ok = 0;
      break;
    }
    if (ok == 1) {
      cout << "YES" << endl;
      return 0;
    }
    for (int flag2 = 1; flag2 < 3; flag2++) {
      ind2 = (flag2 == 1 ? fir[j] : sec[j]);
      ok = 1;
      for (int k = j + 1; k <= m; k++) {
        if (fir[k] == ind || fir[k] == ind2 || sec[k] == ind || sec[k] == ind2)
          continue;
        ok = 0;
        break;
      }
      if (ok == 1) {
        cout << "YES" << endl;
        return 0;
      }
    }
  }
  cout << "NO" << endl;
  return 0;
}
