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 T = 1;
  while (T--) {
    int n, m;
    cin >> n >> m;
    map<int, int> mp;
    vector<pair<int, int> > pp;
    for (int i = 0; i < m; i++) {
      int a, b;
      cin >> a >> b;
      pp.push_back({a, b});
      if (a != b) {
        mp[a]++;
        mp[b]++;
      } else
        mp[a]++;
    }
    vector<pair<int, int> > maxi;
    for (auto x : mp) {
      maxi.push_back({x.second, x.first});
    }
    sort(maxi.rbegin(), maxi.rend());
    int mm = min(4, int((maxi).size()));
    for (int q = 0; q < mm; q++) {
      vector<int> v(m, 0);
      int cnt = 0;
      for (int i = 0; i < m; i++) {
        if (pp[i].first == maxi[q].second || pp[i].second == maxi[q].second) {
          v[i] = 1;
          cnt++;
        }
      }
      cnt = m - cnt;
      if (cnt == 0) {
        cout << "YES";
        exit(0);
      }
      mp.clear();
      for (int i = 0; i < m; i++) {
        if (v[i] == 0) {
          if (pp[i].first != pp[i].second) {
            mp[pp[i].first]++;
            mp[pp[i].second]++;
          } else
            mp[pp[i].first]++;
        }
      }
      for (auto x : mp) {
        if (x.second == cnt) {
          cout << "YES";
          exit(0);
        }
      }
    }
    cout << "NO";
  }
  return 0;
}
