#include <bits/stdc++.h>
using namespace std;
int L, m;
char a[1000];
int pan() {
  int j, i;
  set<char> s;
  for (i = 0; i < L; i++) {
    s.insert(a[i]);
  }
  if (s.size() > m)
    return 1;
  else
    return 0;
}
int main() {
  int j, i, w;
  scanf("%d", &w);
  while (w--) {
    scanf("%s %d", a, &m);
    L = strlen(a);
    i = 0;
    while (pan()) {
      set<char> s;
      for (i = 0; i < L; i++) {
        s.insert(a[i]);
        if (s.size() > m) {
          while (a[i] == '9') i--;
          a[i]++;
          for (j = i + 1; j < L; j++) a[j] = '0';
          break;
        }
      }
    }
    printf("%s\n", a);
  }
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int L, m;
char a[1000];
int pan() {
  int j, i;
  set<char> s;
  for (i = 0; i < L; i++) {
    s.insert(a[i]);
  }
  if (s.size() > m)
    return 1;
  else
    return 0;
}
int main() {
  int j, i, w;
  scanf("%d", &w);
  while (w--) {
    scanf("%s %d", a, &m);
    L = strlen(a);
    i = 0;
    while (pan()) {
      set<char> s;
      for (i = 0; i < L; i++) {
        s.insert(a[i]);
        if (s.size() > m) {
          while (a[i] == '9') i--;
          a[i]++;
          for (j = i + 1; j < L; j++) a[j] = '0';
          break;
        }
      }
    }
    printf("%s\n", a);
  }
  return 0;
}

```