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 mul(long long a, long long b) {
  return (a % 1000000007 * b % 1000000007) % 1000000007;
}
long long add(long long a, long long b) {
  return (a % 1000000007 + b % 1000000007) % 1000000007;
}
long long powftn(long long a, long long b) {
  if (a == 1 || !b) {
    return 1;
  }
  long long tmp = powftn(a, b / 2);
  tmp = mul(tmp, tmp);
  if (b & 1) {
    tmp = mul(tmp, a);
  }
  return tmp;
}
long long modinv(long long a) { return powftn(a, 1000000007 - 2) % 1000000007; }
long long gcd(long long a, long long b) {
  if (!b) {
    return a;
  }
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  if (!a || !b) {
    return 0;
  }
  return (a * b) / gcd(a, b);
}
int i, j, k, n, m, q, t;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  vector<pair<int, int> > v;
  for (i = 0; i < m; i++) {
    int u, v1;
    cin >> u >> v1;
    v.push_back({u, v1});
  }
  vector<int> f(n + 3, 0);
  int a = v[0].first, c1 = 0, c, d, flag = 0, b = v[0].second;
  for (auto u : v) {
    if (u.first == a || u.second == a)
      c1++;
    else {
      f[u.first]++;
      f[u.second]++;
      if (!flag) {
        flag = 1;
        c = u.first;
        d = u.second;
      }
    }
  }
  int rem = m - c1;
  if (!flag || (f[c] == rem || f[d] == rem)) {
    cout << "YES\n";
    return 0;
  }
  fill(f.begin(), f.end(), 0);
  flag = 0;
  c1 = 0;
  for (auto u : v) {
    if (u.first == b || u.second == b)
      c1++;
    else {
      f[u.first]++;
      f[u.second]++;
      if (!flag) {
        flag = 1;
        c = u.first;
        d = u.second;
      }
    }
  }
  rem = m - c1;
  if (!flag || (f[c] == rem || f[d] == rem)) {
    cout << "YES\n";
    return 0;
  }
  cout << "NO\n";
  return 0;
}
