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;
vector<pair<long long, long long>> pairs;
long long n, m;
bool solve(long long x) {
  long long s, ss;
  bool flag = false, flag1 = false;
  for (long long i = 1; i < m; i++) {
    if (pairs[i].first == x || pairs[i].second == x)
      continue;
    else {
      flag = true;
      s = pairs[i].first;
      ss = pairs[i].second;
      break;
    }
  }
  if (!flag) return true;
  flag = false;
  for (long long i = 1; i < m; i++) {
    if (pairs[i].first == x || pairs[i].second == s || pairs[i].first == s ||
        pairs[i].second == x)
      continue;
    else {
      flag = true;
      break;
    }
  }
  for (long long i = 1; i < m; i++) {
    if (pairs[i].first == x || pairs[i].second == ss || pairs[i].first == ss ||
        pairs[i].second == x)
      continue;
    else {
      flag1 = true;
      break;
    }
  }
  if (flag && flag1)
    return false;
  else
    return true;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long f, s;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    cin >> f >> s;
    pairs.push_back({f, s});
  }
  bool flag = solve(pairs[0].first);
  if (flag) {
    cout << "YES";
    return 0;
  }
  flag = solve(pairs[0].second);
  if (flag)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}
