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 mult(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  return (a % p + b % p) % p;
}
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
long long inv_euclid(long long a, long long m = 1000000007) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  vector<pair<long long, long long> > v;
  long long sum = 0, act = 0, k, n, q, f;
  long long m;
  cin >> n >> m;
  v.reserve(n);
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    v.push_back(make_pair(a, b));
  }
  vector<long long> p = {v[0].first, v[0].second};
  for (auto i : p) {
    vector<long long> val(n);
    long long all = 0;
    long long max = 0;
    for (auto j : v) {
      if (j.first != i && j.second != i) {
        val[j.first]++;
        val[j.second]++;
        all++;
        if (val[j.first] > max) {
          max = val[j.first];
        }
        if (val[j.second] > max) {
          max = val[j.second];
        }
      }
    }
    if (all == max) {
      cout << "YES"
           << "\n";
      return 0;
    }
  }
  cout << "NO"
       << "\n";
  return 0;
}
