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;
  cin >> n >> m;
  vector<pair<long long, long long> > mas(m);
  for (auto &i : mas) {
    cin >> i.first >> i.second;
  }
  long long x = mas[0].second;
  set<long long> a;
  vector<pair<long long, long long> > kek;
  for (auto &i : mas) {
    if (i.first != x && i.second != x) {
      a.insert(i.first);
      a.insert(i.second);
      kek.push_back(i);
    }
  }
  for (auto &it : a) {
    bool ok = 1;
    for (auto &i : kek) {
      if (it != i.first && it != i.second) {
        ok = 0;
        break;
      }
    }
    if (ok) {
      cout << "YES";
      return 0;
    }
  }
  x = mas[0].first;
  a.clear();
  kek.clear();
  for (auto &i : mas) {
    if (i.first != x && i.second != x) {
      a.insert(i.first);
      a.insert(i.second);
      kek.push_back(i);
    }
  }
  if (kek.size() == 0) {
    cout << "YES";
    return 0;
  }
  for (auto &it : a) {
    bool ok = 1;
    for (auto &i : kek) {
      if (it != i.first && it != i.second) {
        ok = 0;
        break;
      }
    }
    if (ok) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
  return 0;
}
