description: It is a complicated version of problem F1. The difference between them is the constraints (F1: k ≤ 2, F2: k ≤ 10).
  
  You are given an integer n. Find the minimum integer x such that x ≥ n and the number x is k-beautiful.
  
  A number is called k-beautiful if its decimal representation having no leading zeroes contains no more than k different digits. E.g. if k = 2, the numbers 3434443, 55550, 777 and 21 are k-beautiful whereas the numbers 120, 445435 and 998244353 are not.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
  
  Each test case consists of one line containing two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 10).
  
  Output
  
  For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.
  
  Example
  
  Input
  
  
  6
  2021 3
  177890 2
  34512 3
  724533 4
  998244353 1
  12345678 10
  
  
  Output
  
  
  2021
  181111
  34533
  724542
  999999999
  12345678
solution: #include <bits/stdc++.h>
  const int inf = 2e9;
  int T, n, k, n2, cnt, m, a[12], b[12], v[12];
  long long ans;
  inline long long min(long long x, long long y) { return x < y ? x : y; }
  long long dfs(int x) {
    if (x == 0) return 0;
    int t = std::lower_bound(b + 1, b + m + 1, a[x]) - b;
    if (t > m) return inf;
    if (b[t] == a[x]) {
      int t2 = dfs(x - 1);
      if (t2 != inf) return 1ll * b[t] * v[x] + t2;
      if (t == m) return inf;
      return 1ll * b[t + 1] * v[x] + 1ll * b[1] * (v[x] - 1) / 9;
    }
    return 1ll * b[t] * v[x] + 1ll * b[1] * (v[x] - 1) / 9;
  }
  int main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d%d", &n, &k), ans = inf;
      n2 = n / 10, a[cnt = 1] = n % 10, v[1] = 1;
      while (n2) a[++cnt] = n2 % 10, v[cnt] = v[cnt - 1] * 10, n2 /= 10;
      for (int i = 1, j; i < 1024; ++i)
        if (__builtin_popcount(i) == k) {
          m = 0, j = i;
          while (j) b[++m] = __builtin_ctz(j), j ^= (1 << b[m]);
          ans = min(ans, dfs(cnt));
        }
      printf("%d\n", ans);
    }
    return 0;
  }
  