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() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long r, n;
  while (cin >> r >> n) {
    vector<long long> v;
    map<pair<long long, long long>, long long> mp;
    map<long long, long long> cnt;
    for (long long i = 0; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      if (x > y) swap(x, y);
      mp[make_pair(x, y)]++;
      if (!cnt[x]) v.push_back(x);
      if (!cnt[y]) v.push_back(y);
      cnt[x]++;
      cnt[y]++;
    }
    long long l = v.size();
    long long ok = 0;
    for (long long i = 0; i < 2; i++) {
      for (long long j = i + 1; j < l; j++) {
        long long x = v[i];
        long long y = v[j];
        if (x > y) swap(x, y);
        if (cnt[v[i]] + cnt[v[j]] - mp[make_pair(x, y)] == n) {
          ok = 1;
          break;
        }
      }
    }
    if (ok || l == 2)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}
