...
void execute(int test_number)
{
  cin>>n>>k>>str;
  vector <int> count_char(26, 0);
  for (char c: str) count_char[c - 'a']++;
  string ans = "";
  for (int i = 0; i < min(25, n/k); i++) {
    while (k - ans.size() > count_char[i]) {
      ans.push_back(i + 'a');
    }
  }

  char c = 'a' + min(n / k, 25);
  while (k > ans.size()) {
    ans += c;
  }
  reverse(ans.begin(), ans.end());
  cout << ans << "\n";
}
...

...
ll l, r;

ll bs_sqrt(ll x) {
  ll left = 0, right = 2000000123;
  while (right > left) {
      ll mid = (left + right) / 2;
      
      if (mid * mid > x) right = mid;
      else left = mid + 1;
  }
  return left - 1;
}

// main solution goes here:
void execute(int test_number)
{
  cin >> l >> r;
  ll sql = bs_sqrt(l), sqr = bs_sqrt(r);
  ll ans;
  if (sql == sqr) {
    ans = 0;
    for (int i = 0; i < 3; i++) {
      if (l <= sql * (sql + i) && sql * (sql + i) <= r) ans++;
    }
  } else {
    ans = (sqr - sql - 1) * 3;
    for (int i = 0; i < 3; i++) {
      if (l <= sql * (sql + i) && sql * (sql + i) <= r) ans++;
      if (l <= sqr * (sqr + i) && sqr * (sqr + i) <= r) ans++;
    }
  }
  cout << ans << "\n";
}
...

...
int n;
int x[3], y[3];
int u, v;

pii centralSquare() {
  int a = (x[0] == x[1]) ? x[0] : x[2];
  int b = (y[0] == y[1]) ? y[0] : y[2];
  return {a, b};
}
// main solution goes here:
void execute(int test_number)
{
  cin>>n;
  for (int i=0; i<3; i++) cin>>x[i]>>y[i];
  cin>>u>>v;

  int cx = centralSquare().first, cy = centralSquare().second;

  if ((cx == 1 || cx == n) && (cy == 1 || cy == n)) { // "corner" case, literally
    // the crickets can only reach coordinates within the edges that already contains at least 2 crickets,
    // which contains the centralSquare of the L
    cout << ((u == cx || v == cy) ? "YES\n" : "NO\n");
  } else {
    if ((cx + cy) % 2 == (u + v) % 2) {
      cout << (cx % 2 == u % 2 ? "YES\n" : "NO\n"); 
    } else { // can be prove to always reach, since we have ways to align 2 crickets in the same diagonal as target
      cout << "YES\n"; 
    }
  }
}
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