#include <bits/stdc++.h>
using namespace std;
long long int modpwr(long long int x, long long int y, long long int m) {
  x %= m;
  if (x == 0) return 0;
  long long int ans = 1;
  while (y) {
    if (y % 2) {
      ans = (ans * x) % m;
    }
    x = (x * x) % m;
    y = y / 2;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    string n;
    int k;
    cin >> n >> 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();
      for (int i = 0; i < n.size(); i++) {
        s.insert(n[i]);
        if (s.size() > k) {
          int j = i;
          while (n[j] == '9') j--;
          n[j]++;
          j++;
          for (; j < n.size(); j++) n[j] = '0';
          break;
        }
      }
    }
  }
}

### Prompt

Generate a cpp solution to 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;
long long int modpwr(long long int x, long long int y, long long int m) {
  x %= m;
  if (x == 0) return 0;
  long long int ans = 1;
  while (y) {
    if (y % 2) {
      ans = (ans * x) % m;
    }
    x = (x * x) % m;
    y = y / 2;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    string n;
    int k;
    cin >> n >> 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();
      for (int i = 0; i < n.size(); i++) {
        s.insert(n[i]);
        if (s.size() > k) {
          int j = i;
          while (n[j] == '9') j--;
          n[j]++;
          j++;
          for (; j < n.size(); j++) n[j] = '0';
          break;
        }
      }
    }
  }
}

```