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 unsigned long long N = 300005;
int n, q, m;
set<pair<int, int> > s;
pair<int, int> v[N];
bool ok(int x, int y) {
  for (int i = 1; i < m; ++i) {
    int x1, y1;
    tie(x1, y1) = v[i];
    if (x1 != x && y1 != x && x1 != y && y1 != y) {
      return false;
    }
  }
  return true;
}
void solve() {
  cin >> n >> m;
  int x, y;
  for (int i = 0; i < m; ++i) {
    cin >> x >> y;
    v[i] = {x, y};
  }
  for (auto x : {v[0].first, v[0].second}) {
    if (ok(x, -1)) {
      cout << "YES\n";
      return;
    }
    for (int i = 1; i < m; ++i) {
      int x1, y1;
      tie(x1, y1) = v[i];
      if (x != x1 && x != y1) {
        if (ok(x, x1) || ok(x, y1)) {
          cout << "YES\n";
          return;
        }
        break;
      }
    }
  }
  cout << "NO\n";
}
int main(int argc, char* argv[]) {
  solve();
  return 0;
}
