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>
  using namespace std;
  const int maxn = 1e6 + 5;
  const int mod = 1e9 + 7;
  int digit[maxn];
  int dp[10][2][1 << 10];
  int n, k;
  int g[1 << 10];
  int dfs(int len, bool lim, int s) {
    if (len == 0) return g[s] <= k;
    int &x = dp[len][lim][s];
    if (~x) return x;
    int low = lim ? digit[len] : 0;
    int ans = false;
    for (int i = low; i <= 9; i++) {
      ans |= dfs(len - 1, lim && i == low, s | (1 << i));
      if (ans) break;
    }
    return x = ans;
  }
  long long ksm(long long a, long long b) {
    long long ans = 1, base = a;
    while (b) {
      if (b & 1) ans = ans * base % mod;
      b >>= 1;
      base = base * base % mod;
    }
    return ans;
  }
  long long getans(int len, int lim, int s) {
    if (len == 0) return 0;
    int low = lim ? digit[len] : 0;
    for (int i = low; i <= 9; i++) {
      bool ok = dfs(len - 1, lim && i == low, s | (1 << i));
      if (ok) {
        return i * ksm(10, len - 1) +
               getans(len - 1, lim && i == low, s | (1 << i));
      }
    }
    return -1;
  }
  long long solve() {
    int k = 0;
    while (n) {
      digit[++k] = n % 10;
      n /= 10;
    }
    memset(dp, -1, sizeof dp);
    dfs(k, true, 0);
    return getans(k, true, 0);
  }
  int main() {
    ios::sync_with_stdio(false);
    g[0] = 0;
    for (int i = 1; i < 1 << 10; i++) {
      g[i] = g[i >> 1] + (i & 1);
    }
    int t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      cout << solve() << endl;
    }
    return 0;
  }
  