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 long long N = 1e6 + 10;
  long long n, k, dp[12][(1 << 10)][3][12], a[12], cnt, res = 0;
  bool check(long long msk, long long cur) {
    long long cnt = 0;
    while (msk) {
      cnt += (msk & 1);
      msk >>= 1;
    }
    return (cnt <= cur);
  }
  void calc(long long i, long long msk, long long c, long long cur) {
    dp[i][msk][c][cur] = 0;
    if (i == 0) {
      dp[i][msk][c][cur] = check(msk, cur);
      return;
    }
    for (long long j = 0; j <= 9; j++) {
      long long msk2;
      if (msk == 0 && j == 0)
        msk2 = 0;
      else
        msk2 = msk | (1 << j);
      if (dp[i - 1][msk2][c][cur] == -1) calc(i - 1, msk2, c, cur);
      dp[i][msk][c][cur] |= dp[i - 1][msk2][c][cur];
    }
  }
  void calc2(long long i, long long msk, long long c, long long cur) {
    dp[i][msk][c][cur] = 0;
    if (i == 0) {
      dp[i][msk][c][cur] = check(msk, cur);
      return;
    }
    for (long long j = a[i]; j <= 9; j++) {
      long long msk2;
      if (msk == 0 && j == 0)
        msk2 = 0;
      else
        msk2 = msk | (1 << j);
      if (a[i] == j && dp[i - 1][msk2][c & (a[i] == j)][cur] == -1) {
        calc2(i - 1, msk2, c & (a[i] == j), cur);
      }
      dp[i][msk][c][cur] |= dp[i - 1][msk2][c & (a[i] == j)][cur];
    }
  }
  void print_result(long long i, long long msk, long long c) {
    if (i == 0) {
      return;
    }
    if (c == 0)
      for (long long j = 0; j <= 9; j++) {
        long long msk2;
        if (msk == 0 && j == 0)
          msk2 = 0;
        else
          msk2 = msk | (1 << j);
        if (dp[i - 1][msk2][c][k] == 1) {
          res *= 10;
          res += j;
          print_result(i - 1, msk2, c);
          break;
        }
      }
    else
      for (long long j = a[i]; j <= 9; j++) {
        long long msk2;
        if (msk == 0 && j == 0)
          msk2 = 0;
        else
          msk2 = msk | (1 << j);
        if (dp[i - 1][msk2][c & (a[i] == j)][k] == 1) {
          res *= 10;
          res += j;
          print_result(i - 1, msk2, c & (a[i] == j));
          break;
        }
      }
  }
  void solve() {
    cin >> n >> k;
    for (long long i = 1; i <= 10; i++) a[i] = 0;
    long long j = 0;
    while (n > 0) {
      a[++j] = n % 10;
      n /= 10;
    }
    calc2(j + 1, 0LL, 1LL, k);
    print_result(j + 1, 0LL, 1LL);
    cout << res << "\n";
    res = 0;
    for (long long i = 0; i <= j + 1; i++)
      for (long long msk = 0; msk <= (1 << 10); msk++) dp[i][msk][1][k] = -1;
  }
  int main() {
    memset(dp, -1, sizeof(dp));
    ios_base::sync_with_stdio(NULL);
    cin.tie(0);
    cout.tie(0);
    for (long long i = 1; i <= 10; i++) calc(10LL, 0LL, 0LL, i);
    long long t = 1;
    cin >> t;
    while (t--) solve();
    cerr << "\n" << (float)clock() / CLOCKS_PER_SEC * 1000 << " ms";
  }
  