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;
int n, m;
int x[300000];
int y[300000];
int main() {
  cin >> n >> m;
  set<int> nodes;
  for (int i = 0; i < m; ++i) {
    cin >> x[i] >> y[i];
    if (nodes.size() >= 4) continue;
    auto itx = nodes.find(x[i]);
    auto ity = nodes.find(y[i]);
    if (itx == nodes.end() && ity == nodes.end()) {
      nodes.insert(x[i]);
      nodes.insert(y[i]);
    }
  }
  if (nodes.size() <= 3) {
    cout << "YES" << endl;
  } else {
    int v[4];
    auto it = nodes.begin();
    for (int i = 0; i < 4; ++i) {
      v[i] = *it;
      it++;
    }
    bool yes = false;
    for (int i = 0; i < 4; ++i) {
      for (int j = i + 1; j < 4; ++j) {
        bool valid = true;
        for (int k = 0; k < m; ++k) {
          if (x[k] != v[i] && y[k] != v[i] && x[k] != v[j] && y[k] != v[j]) {
            valid = false;
            break;
          }
        }
        if (valid) yes = true;
      }
    }
    cout << (yes ? "YES" : "NO") << endl;
  }
  return 0;
}
