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 i, t, n, j, m, g;
  cin >> n >> m;
  map<int, int> mp;
  map<pair<int, int>, int> mpp;
  for (i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    mp[a]++;
    mp[b]++;
    mpp[{a, b}]++;
  }
  vector<pair<int, int>> v;
  for (auto it = mp.begin(); it != mp.end(); it++) {
    v.push_back({it->second, it->first});
  }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  int f = 0;
  int jj = v.size();
  for (i = 0; i < min(4, jj); ++i) {
    for (j = i + 1; j < jj; ++j) {
      if (v[i].first + v[j].first - mpp[{v[i].second, v[j].second}] -
              mpp[{v[j].second, v[i].second}] >=
          m) {
        f = 1;
        break;
      }
    }
    if (f == 1) {
      break;
    }
  }
  if (f == 0) {
    cout << "NO"
         << "\n";
  } else {
    cout << "YES"
         << "\n";
  }
  return 0;
}
