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>
  #pragma warning(disable : 4996)
  int T, n, k, a, b, c;
  char care[100];
  char str[100];
  int dif(char* str) {
    int is = 0;
    int n = strlen(str);
    int ans = 0;
    int bulk[10];
    for (int i = 0; i < 10; i++) {
      bulk[i] = 0;
    }
    for (is; is < n; is++) {
      if (bulk[str[is] - '0'] == 0) ans++;
      bulk[str[is] - '0']++;
    }
    return ans;
  }
  bool hasbig(int* a, char b) {
    for (int i = b + 1; i < 10; i++) {
      if (a[i]) {
        return true;
      }
    }
    return false;
  }
  char nextbig(int* a, char b) {
    for (int i = b + 1; i < 10; i++) {
      if (a[i]) {
        return i;
      }
    }
    return 10;
  }
  char least(int* a) {
    for (int i = 0; i < 10; i++) {
      if (a[i]) {
        return i;
      }
    }
    return 10;
  }
  bool full(char* a, char* b, int k) {
    int is = 0;
    int n = strlen(a);
    b[n] = '\0';
    int bulk[10];
    for (int i = 0; i < 10; i++) {
      bulk[i] = 0;
    }
    int omit = 0;
    for (is; is < n; is++) {
      b[is] = a[is];
      if (bulk[a[is] - '0'] == 0) omit++;
      bulk[a[is] - '0']++;
      if (omit == k) break;
    }
    for (is; is < n; is++) {
      if (bulk[a[is] - '0']) {
        b[is] = a[is];
      } else {
        break;
      }
    }
    if (is == n) {
      return true;
    }
    if (hasbig(bulk, a[is] - '0')) {
      b[is] = nextbig(bulk, a[is] - '0') + '0';
      for (int i = is + 1; i < n; i++) {
        b[i] = least(bulk) + '0';
      }
    } else {
      return false;
    }
  }
  void take(char* a, char* b, int c) {
    int is = 0;
    int n = strlen(a);
    b[n] = '\0';
    int bulk[10];
    for (int i = 0; i < 10; i++) {
      bulk[i] = 0;
    }
    int omit = 0;
    for (is; is < n; is++) {
      b[is] = a[is];
      if (bulk[a[is] - '0'] == 0) omit++;
      bulk[a[is] - '0']++;
      if (omit == k + 1) break;
    }
    bulk[b[is] - '0'] = 0;
    is--;
    while (bulk[b[is] - '0'] != 1) {
      if (hasbig(bulk, b[is] - '0')) {
        b[is] = nextbig(bulk, b[is] - '0') + '0';
        for (is++; is < n; is++) {
          b[is] = least(bulk) + '0';
        }
        return;
      } else {
        bulk[b[is] - '0']--;
        is--;
      }
    }
    if (b[is] == '9') {
      if (bulk[b[is - 1] - '0'] == 1) {
        bulk[b[is - 1] - '0']--;
        b[is - 1]++;
        bulk[b[is - 1] - '0']++;
        for (is; is < n; is++) {
          b[is] = '0';
        }
        return;
      } else {
        bulk[b[is - 1] - '0']--;
        b[is - 1]++;
        bulk[b[is - 1] - '0']++;
        for (is; is < n; is++) {
          b[is] = least(bulk) + '0';
        }
        return;
      }
    } else {
      bulk[b[is] - '0']--;
      b[is]++;
      bulk[b[is] - '0']++;
      if (bulk[b[is] - '0'] == 1) {
        for (is++; is < n; is++) {
          b[is] = least(bulk) + '0';
        }
      } else {
        for (is++; is < n; is++) {
          b[is] = '0';
        }
      }
    }
  }
  int main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d%d", &n, &k);
      sprintf(care, "%d", n);
      a = dif(care);
      if (a <= k) {
        printf("%d\n", n);
        continue;
      }
      if (full(care, str, k)) {
        printf("%s\n", str);
        continue;
      } else {
        take(care, str, k);
        printf("%s\n", str);
      }
    }
  }
  