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() {
  int n, m;
  cin >> n >> m;
  int f = 0;
  int a[m][2];
  map<int, int> mp;
  for (int i = 0; i < m; i++) {
    cin >> a[i][0] >> a[i][1];
    mp[a[i][0]]++;
    mp[a[i][1]]++;
  }
  map<int, int>::iterator itr;
  vector<int> l;
  int p = 0, q = 0;
  for (itr = mp.begin(); itr != mp.end(); ++itr) {
    if (itr->second >= m / 2) {
      l.push_back(itr->first);
    }
  }
  int count = 0;
  map<int, int> mp1;
  int c = 0;
  for (int j = 0; j < l.size(); j++) {
    p = l[j];
    c = 0;
    for (int i = 0; i < m; i++) {
      if (p == a[i][0] || p == a[i][1]) {
        c++;
      } else {
        mp1[a[i][0]]++;
        mp1[a[i][1]]++;
      }
    }
    for (itr = mp1.begin(); itr != mp1.end(); ++itr) {
      if ((c + itr->second) == m) {
        f = 1;
        break;
      }
    }
    if (c == m) {
      f = 1;
      break;
    }
    if (f == 1) break;
    mp1.clear();
  }
  if (f == 1)
    cout << "YES";
  else
    cout << "NO";
}
