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;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.second < b.second) return 1;
  if (a.second == b.second && a.first < b.first) return 1;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m;
  cin >> n >> m;
  pair<long long, long long> arr[m];
  for (long long i = 0; i < m; i++) cin >> arr[i].first >> arr[i].second;
  long long val1 = arr[0].first;
  long long val2 = arr[0].second;
  for (long long i = 0; i < m; i++) {
    if (arr[i].first != arr[i].second) {
      val1 = arr[i].first;
      val2 = arr[i].second;
      break;
    }
  }
  vector<pair<long long, long long>> v1, v2;
  for (long long i = 0; i < m; i++) {
    if (arr[i].first != val1 && arr[i].second != val1) v1.push_back(arr[i]);
    if (arr[i].first != val2 && arr[i].second != val2) v2.push_back(arr[i]);
  }
  unordered_map<long long, long long> ma1, ma2;
  for (long long i = 0; i < v1.size(); i++) {
    if (v1[i].first == v1[i].second)
      ma1[v1[i].first]++;
    else {
      ma1[v1[i].first]++;
      ma1[v1[i].second]++;
    }
  }
  for (long long i = 0; i < v2.size(); i++) {
    if (v2[i].first == v2[i].second)
      ma2[v2[i].first]++;
    else {
      ma2[v2[i].first]++;
      ma2[v2[i].second]++;
    }
  }
  bool pos1 = 0, pos2 = 0;
  if (ma1.size() == 0) pos1 = 1;
  if (ma2.size() == 0) pos2 = 1;
  for (auto it = ma1.begin(); it != ma1.end(); it++) {
    if (it->second == v1.size()) {
      pos1 = 1;
      break;
    }
  }
  for (auto it = ma2.begin(); it != ma2.end(); it++) {
    if (it->second == v2.size()) {
      pos2 = 1;
      break;
    }
  }
  if (pos1 == 1 || pos2 == 1)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}
