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 MAXN = 3e5 + 13;
int a[MAXN], b[MAXN];
bool check(vector<pair<int, int> > &A, int x) {
  vector<int> el;
  int i = 0;
  for (; i < int((A).size()); i++) {
    if (A[i].first == x || A[i].second == x) continue;
    el.push_back(i);
  }
  if (int((el).size()) == 0) return true;
  unordered_map<int, int> cnt;
  for (int e : el) {
    cnt[A[e].first]++;
    cnt[A[e].second]++;
    if (cnt[A[e].first] == int((el).size()) ||
        cnt[A[e].second] == int((el).size()))
      return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m, x, y;
  while (cin >> n >> m) {
    vector<pair<int, int> > A(m);
    for (__typeof(0) i = (0); i < (m); i++) {
      cin >> A[i].first >> A[i].second;
    }
    if (check(A, A[0].first) || check(A, A[0].second)) {
      cout << "YES" << '\n';
    } else {
      cout << "NO" << '\n';
    }
  }
  return 0;
}
