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;
long long n, m, a[300010], b[300010], cnt[300010];
bool solve(long long t) {
  for (int i = 1; i <= n; i++) cnt[i] = 0;
  long long all = 0;
  for (int i = 0; i < m; i++) {
    if (a[i] == t || b[i] == t) continue;
    all++;
    cnt[a[i]]++, cnt[b[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    if (cnt[i] == all) return 1;
  }
  return 0;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a[i] >> b[i];
  }
  bool ans = 0;
  ans |= solve(a[0]);
  ans |= solve(b[0]);
  cout << (ans ? "YES" : "NO") << endl;
  return 0;
}
