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;
signed main() {
  long long n, m, a, b, ans2 = -1, ans3 = -1, r = 0, t = 0, c = 0, ans1;
  cin >> n >> m;
  set<pair<long long, long long> > s;
  for (long long i = 0; i < m; i++) {
    cin >> a >> b;
    s.insert({min(a, b), max(a, b)});
  }
  if (s.size() == 1) {
    cout << "YES";
    return 0;
  }
  map<long long, long long> mp;
  for (auto x : s) {
    mp[x.first]++;
    mp[x.second]++;
  }
  long long mx = -1;
  for (auto x : mp)
    if (x.second > mx) {
      mx = x.second;
      ans1 = x.first;
    }
  for (auto x : s) {
    if (x.first == ans1 || x.second == ans1)
      continue;
    else {
      ans2 = x.first;
      ans3 = x.second;
      break;
    }
  }
  if (ans2 > 0 && ans3 > 0) {
    for (auto x : s)
      if (x.first == ans1 || x.second == ans1)
        continue;
      else {
        c++;
        if (x.first == ans2 || x.second == ans2) r++;
        if (x.first == ans3 || x.second == ans3) t++;
      }
    if (r == c || t == c) {
      cout << "YES";
      return 0;
    }
  } else {
    cout << "YES";
    return 0;
  }
  cout << "NO";
}
