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;
pair<int, int> a[3000005];
int n, m;
int x, y;
inline bool g(int second, int k) {
  for (int i = second + 1; i <= m; i++) {
    if (a[i].first != k && a[i].second != k && a[i].first != x &&
        a[i].second != x)
      return 0;
  }
  return 1;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  int i;
  cin >> n >> m;
  for (i = 1; i <= m; i++) {
    cin >> a[i].first >> a[i].second;
  }
  x = a[1].first;
  for (i = 2; i <= m; i++) {
    if (a[i].first != x && a[i].second != x) {
      int k = max(g(i, a[i].first), g(i, a[i].second));
      if (k == 1) {
        cout << "YES";
        return 0;
      } else {
        break;
      }
    }
  }
  if (i > m) {
    cout << "YES";
    return 0;
  }
  x = a[1].second;
  for (i = 2; i <= m; i++) {
    if (a[i].first != x && a[i].second != x) {
      int k = max(g(i, a[i].first), g(i, a[i].second));
      if (k == 1) {
        cout << "YES";
        return 0;
      } else
        break;
    }
  }
  if (i > m) {
    cout << "YES";
  } else
    cout << "NO";
  return 0;
}
