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 fn(vector<pair<int, int>>& v, int n, int m, int x) {
  bool a[m];
  int i, p, q, cp = 0, cq = 0, cx = 0;
  for (i = 0; i < m; i++) {
    if (v[i].first == x || v[i].second == x) {
      a[i] = true;
      cx++;
    } else
      a[i] = false;
    if (v[i].first != x && v[i].second != x) {
      p = v[i].first;
      q = v[i].second;
    }
  }
  for (i = 0; i < m; i++) {
    if (a[i] == false) {
      if (v[i].first == p || v[i].second == p) cp++;
      if (v[i].first == q || v[i].second == q) cq++;
    }
  }
  int cy = max(cp, cq);
  if (cx + cy >= m)
    return 1;
  else
    return 0;
}
int main() {
  long long int n, m, a, b, x, y, ans = INT_MIN, i;
  cin >> n >> m;
  vector<pair<int, int>> v;
  for (i = 0; i < m; i++) {
    cin >> a >> b;
    v.push_back(make_pair(a, b));
  }
  int ans1 = fn(v, n, m, v[0].first);
  int ans2 = fn(v, n, m, v[0].second);
  if (ans1 == 1 || ans2 == 1)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}
