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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
bool check_two(long long first, long long second, vector<long long> &a,
               vector<long long> &b) {
  for (long long i = 0; i < (long long)(a).size(); i++) {
    if (first == a[i] || first == b[i]) {
      continue;
    } else if (second == a[i] || second == b[i]) {
      continue;
    }
    return false;
  }
  return true;
}
bool check(long long first, vector<long long> &a, vector<long long> &b) {
  map<long long, long long> mp;
  for (long long i = 0; i < (long long)(a).size(); i++) {
    if (first == a[i] || first == b[i]) {
      continue;
    }
    return (check_two(first, a[i], a, b) || check_two(first, b[i], a, b));
  }
  return true;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<long long> a(m), b(m);
  for (long long i = 0; i < m; i++) {
    cin >> a[i] >> b[i];
  }
  if (check(a[0], a, b) || check(b[0], a, b)) {
    cout << "YES\n";
  } else {
    cout << "NO\n";
  }
  return 0;
}
