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 co[500005] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, x, y;
  cin >> n >> m;
  int maxi = 0, in = -1;
  map<pair<int, int>, int> ma;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    ma[{min(x, y), max(x, y)}]++;
    co[x]++;
    co[y]++;
    if (co[x] > maxi) {
      maxi = co[x];
      in = x;
    }
    if (co[y] > maxi) {
      maxi = co[y];
      in = y;
    }
  }
  int f = 0;
  vector<pair<int, int> > v;
  for (int i = 1; i <= n; i++) {
    if (co[i] > 0) {
      v.push_back({co[i], i});
    }
  }
  sort(v.begin(), v.end());
  for (int i = v.size() - 1; i >= 0; i--) {
    for (int j = i - 1; j >= 0; j--) {
      if (v[j].first + v[i].first -
              ma[{min(v[j].second, v[i].second),
                  max(v[j].second, v[i].second)}] >=
          m) {
        f = 1;
        break;
      }
      if (v[j].first < m - v[i].first) break;
    }
    if (f) break;
    if (v[i].first < (m / 2)) break;
  }
  if (f)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}
