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 ll = long long int;
using namespace std;
vector<pair<int, int> > vp;
int n, m;
int ok(int x, int i, int y) {
  for (; i < m; ++i) {
    if (vp[i].first == x || vp[i].second == x || vp[i].second == y ||
        vp[i].first == y)
      continue;
    return i;
  }
  return m;
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    vp.push_back({x, y});
  }
  int x = vp[0].first, y = vp[0].second;
  int first_it = ok(x, 0, 0), second_it = ok(y, 0, 0);
  if (first_it == m || second_it == m) {
    cout << "YES";
    return;
  } else {
    int x1 = vp[first_it].first, y1 = vp[first_it].second;
    int id_1 = ok(x1, first_it, x), id_2 = ok(y1, first_it, x);
    if (id_1 == m || id_2 == m) {
      cout << "YES";
      return;
    }
    x1 = vp[second_it].first, y1 = vp[second_it].second;
    int id1 = ok(x1, second_it, y), id2 = ok(y1, second_it, y);
    if (id1 == m || id2 == m) {
      cout << "YES";
      return;
    }
  }
  cout << "NO"
       << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
