#include <bits/stdc++.h>
using namespace std;
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string n;
    int k;
    cin >> n;
    cin >> k;
    while (true) {
      set<char> s;
      for (int i = 0; i < n.size(); i++) s.insert(n[i]);
      if (s.size() <= k) {
        cout << n << "\n";
        break;
      }
      s.clear();
      int ptr = 0;
      while (true) {
        s.insert(n[ptr]);
        if (s.size() > k) {
          while (n[ptr] == '9') ptr--;
          n[ptr]++;
          for (int i = ptr + 1; i < n.size(); i++) n[i] = '0';
          break;
        }
        ptr++;
      }
    }
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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 dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string n;
    int k;
    cin >> n;
    cin >> k;
    while (true) {
      set<char> s;
      for (int i = 0; i < n.size(); i++) s.insert(n[i]);
      if (s.size() <= k) {
        cout << n << "\n";
        break;
      }
      s.clear();
      int ptr = 0;
      while (true) {
        s.insert(n[ptr]);
        if (s.size() > k) {
          while (n[ptr] == '9') ptr--;
          n[ptr]++;
          for (int i = ptr + 1; i < n.size(); i++) n[i] = '0';
          break;
        }
        ptr++;
      }
    }
  }
}

```