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 main() {
  long long n, m;
  cin >> n >> m;
  long long a[m], b[m];
  for (int i = 0; i < m; i++) {
    cin >> a[i] >> b[i];
  }
  long long t = a[0];
  bool ans1 = true, ans2 = true, ans3 = true, ans4 = true;
  int i = 0;
  while (a[i] == t || b[i] == t) {
    i++;
    if (i == m) {
      ans1 = true;
      break;
    }
  }
  if (i < m) {
    long long t1 = a[i];
    for (long long j = i; j < m; j++) {
      if (a[j] != t && b[j] != t && a[j] != t1 && b[j] != t1) {
        ans1 = false;
        break;
      }
    }
    long long t2 = b[i];
    for (long long j = i; j < m; j++) {
      if (a[j] != t && b[j] != t && a[j] != t2 && b[j] != t2) {
        ans2 = false;
        break;
      }
    }
  }
  i = 0;
  t = b[0];
  while (a[i] == t || b[i] == t) {
    i++;
    if (i == m) {
      ans1 = true;
      break;
    }
  }
  if (i < m) {
    long long t1 = a[i];
    for (long long j = i; j < m; j++) {
      if (a[j] != t && b[j] != t && a[j] != t1 && b[j] != t1) {
        ans3 = false;
        break;
      }
    }
    long long t2 = b[i];
    for (long long j = i; j < m; j++) {
      if (a[j] != t && b[j] != t && a[j] != t2 && b[j] != t2) {
        ans4 = false;
        break;
      }
    }
  }
  bool ans = ans1 | ans2 | ans3 | ans4;
  if (ans) {
    cout << "YES";
  } else {
    cout << "NO";
  }
  return 0;
}
