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() {
  long long n, m, f1, f2, ff1, ss1, ff2, ss2, flag = 0;
  cin >> n >> m;
  vector<pair<long long, long long>> v(m);
  for (long long i = 0; i < m; i++) {
    cin >> v[i].first >> v[i].second;
  }
  f1 = v[0].first;
  f2 = v[0].second;
  for (long long i = 1; i < m; i++) {
    if (v[i].first != f1 && v[i].second != f1) {
      ff1 = v[i].first;
      ss1 = v[i].second;
    }
  }
  for (long long i = 1; i < m; i++) {
    if (v[i].first != f2 && v[i].second != f2) {
      ff2 = v[i].first;
      ss2 = v[i].second;
    }
  }
  for (long long i = 0; i < m; i++) {
    if ((v[i].first != f1 && v[i].second != f1) &&
        (v[i].first != ff1 && v[i].second != ff1)) {
      flag++;
      break;
    }
  }
  if (flag == 1) {
    for (long long i = 0; i < m; i++) {
      if ((v[i].first != f1 && v[i].second != f1) &&
          (v[i].first != ss1 && v[i].second != ss1)) {
        flag++;
        break;
      }
    }
    if (flag == 2) {
      for (long long i = 0; i < m; i++) {
        if ((v[i].first != f2 && v[i].second != f2) &&
            (v[i].first != ff2 && v[i].second != ff2)) {
          flag++;
          break;
        }
      }
    }
    if (flag == 3) {
      for (long long i = 0; i < m; i++) {
        if ((v[i].first != f2 && v[i].second != f2) &&
            (v[i].first != ss2 && v[i].second != ss2)) {
          flag++;
          break;
        }
      }
    }
  }
  if (flag == 4)
    cout << "NO" << endl;
  else
    cout << "YES" << endl;
  return 0;
}
