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 N = 3e5 + 15;
int n, m, a[N][2];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    cin >> a[i][0] >> a[i][1];
  }
  for (int i = 0; i < 2; ++i) {
    int cnt = 0;
    vector<int> t;
    for (int j = 1; j < m; ++j)
      if (a[j][0] != a[0][i] && a[j][1] != a[0][i]) {
        t.push_back(a[j][0]);
        t.push_back(a[j][1]);
        ++cnt;
      }
    if (cnt == 0) {
      cout << "YES";
      return 0;
    }
    sort(t.begin(), t.end());
    for (int x : t) {
      int len = upper_bound(t.begin(), t.end(), x) -
                lower_bound(t.begin(), t.end(), x);
      if (len == cnt) {
        cout << "YES";
        return 0;
      }
    }
  }
  cout << "NO";
  return 0;
}
