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 MOD = 1e9 + 7;
const long long N = 3e5 + 10;
long long gcd(long long n, long long m) {
  if (n == 0)
    return m;
  else
    return gcd(m % n, n);
}
long long a[N], b[N];
map<long long, long long> mp, mp1, mp2;
int32_t main() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= m; i++) {
    cin >> a[i] >> b[i];
    mp[a[i]]++;
    mp[b[i]]++;
  }
  long long k = mp[a[1]];
  long long k1 = mp[b[1]];
  if (k == m || k1 == m) {
    cout << "YES" << endl;
    return 0;
  }
  for (long long i = 1; i <= m; i++) {
    if (a[i] != a[1] && b[i] != a[1]) {
      mp1[a[i]]++;
      mp1[b[i]]++;
    }
    if (a[i] != b[1] && b[i] != b[1]) {
      mp2[a[i]]++;
      mp2[b[i]]++;
    }
  }
  for (auto it : mp1) {
    if (it.second == m - k) {
      cout << "YES" << endl;
      return 0;
    }
  }
  for (auto it : mp2) {
    if (it.second == m - k1) {
      cout << "YES" << endl;
      return 0;
    }
  }
  cout << "NO" << endl;
}
