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;
const int INF = 1e9 + 7;
int N = 1e6 * 5;
long double EPS = 1 / 1e18;
map<int, int> m;
map<pair<int, int>, int> m1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m2, i;
  cin >> n >> m2;
  pair<int, int> a[400000];
  for (i = 0; i < m2; i++) {
    cin >> a[i].first >> a[i].second;
    m[a[i].first]++;
    m[a[i].second]++;
    if (a[i].first < a[i].second) swap(a[i].first, a[i].second);
    m1[a[i]]++;
  }
  int x = a[0].first, y = a[0].second;
  for (i = 1; i <= n; i++) {
    if (m[i] == 0 || i == x) continue;
    if (m[i] + m[x] - m1[make_pair(max(i, x), min(i, x))] == m2) {
      cout << "YES";
      return 0;
    }
  }
  for (i = 1; i <= n; i++) {
    if (m[i] == 0 || i == y) continue;
    if (m[i] + m[y] - m1[make_pair(max(i, y), min(i, y))] == m2) {
      cout << "YES";
      return 0;
    }
  }
  cout << "NO";
}
