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 flag;
int sum, mini = INT_MAX, maxi, cnt, ans, n, m;
vector<pair<int, int> > p;
int find(int x, int y) {
  for (int i = 0; i < m; i++) {
    if (y != p[i].first && y != p[i].second)
      if (x != p[i].first && x != p[i].second) {
        return 0;
      }
  }
  return 1;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    p.push_back(make_pair(x, y));
  }
  int a[2];
  a[0] = p[0].first;
  a[1] = p[0].second;
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < m; j++) {
      if (a[i] != p[j].first && a[i] != p[j].second) {
        if (find(p[j].first, a[i])) {
          cout << "YES\n";
          return 0;
        } else if (find(p[j].second, a[i])) {
          cout << "YES\n";
          return 0;
        }
        break;
      } else if (j == m - 1) {
        cout << "YES\n";
        return 0;
      }
    }
  }
  cout << "NO\n";
  return 0;
}
