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;
using ll = long long;
int n, m;
map<pair<int, int>, int> cnt;
vector<pair<int, int>> oc;
map<int, vector<pair<int, int>>> oks;
int getHighest(int x) {
  for (int i = 0; i < oc.size(); i++) {
    if (oc[i].second == x || cnt.count(make_pair(x, oc[i].second))) {
      continue;
    }
    return oc[i].first;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  map<int, int> occ;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    cnt[make_pair(a, b)]++;
    cnt[make_pair(b, a)]++;
    occ[a]++;
    occ[b]++;
  }
  for (auto& p : cnt) {
    pair<int, int> q = p.first;
    int num = p.second;
    oks[q.first].push_back({q.second, num});
  }
  for (auto& p : occ) {
    oc.push_back({p.second, p.first});
  }
  sort(oc.begin(), oc.end(), greater<pair<int, int>>());
  for (int i = 1; i <= n; i++) {
    if (!occ.count(i)) continue;
    int maxans = occ[i] + getHighest(i);
    for (auto& p : oks[i]) {
      maxans = max(maxans, occ[i] + occ[p.first] - p.second);
    }
    if (maxans == m) {
      cout << "YES\n";
      return 0;
    }
  }
  cout << "NO\n";
}
