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;
const long long prime = 1e9 + 7;
const long long maxN = 1e5 + 1;
const long double pi = 3.1415926536;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, a, b;
  cin >> n >> m;
  vector<pair<long long, long long>> pairs(m);
  for (long long i = (long long)(0); i <= (long long)(m - 1); i++) {
    cin >> a >> b;
    pairs[i] = make_pair(a, b);
  }
  vector<long long> x = {pairs[0].first, pairs[0].second};
  for (auto i : x) {
    vector<long long> cnt(n + 1, 0);
    long long no_x = 0;
    for (auto &j : pairs) {
      if ((j.first != i) && (j.second != i)) {
        cnt[j.first]++;
        cnt[j.second]++;
        no_x++;
      }
    }
    if (*max_element(cnt.begin(), cnt.end()) == no_x) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
  return 0;
}
