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 F[200005];
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y % 2 != 0) res = res * x % p;
    y = y / 2;
    x = x * x % p;
  }
  return res;
}
long long ncr(long long n, long long r, long long p) {
  if (r > n) return 0;
  long long res = F[n];
  res = (res * power(F[r], p - 2ll, p)) % p;
  res = (res * power(F[n - r], p - 2ll, p)) % p;
  return res;
}
long long gcd(long long a, long long b) {
  if (a > b) swap(a, b);
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, long long>> a(m);
  for (long long i = 0; i < m; i++) cin >> a[i].first >> a[i].second;
  bool flag = false, f1 = true, f2 = true;
  long long x = a[0].first;
  long long cnt = 0;
  map<long long, long long> mp;
  for (long long i = 0; i < m; i++) {
    if (a[i].first != x && a[i].second != x) {
      f1 = false;
      cnt++;
      mp[a[i].first]++;
      mp[a[i].second]++;
    }
  }
  for (auto x : mp) {
    if (x.second == cnt) {
      flag = true;
    }
  }
  mp.clear();
  x = a[0].second;
  cnt = 0;
  for (long long i = 0; i < m; i++) {
    if (a[i].first != x && a[i].second != x) {
      f2 = false;
      cnt++;
      mp[a[i].first]++;
      mp[a[i].second]++;
    }
  }
  for (auto x : mp) {
    if (x.second == cnt) {
      flag = true;
    }
  }
  if (flag || f1 || f2)
    cout << "YES\n";
  else
    cout << "NO\n";
}
