#include <bits/stdc++.h>
using namespace std;
void solve() {
  string n, v;
  int k;
  cin >> n >> k;
  while (true) {
    set<char> s;
    int len = n.length();
    for (int i = 0; i < len; i++) {
      s.insert(n[i]);
      if (s.size() > k) {
        while (n[i] == '9') i--;
        n[i]++;
        for (int j = i + 1; j < n.size(); j++) n[j] = '0';
        break;
      }
      if (i == len - 1) {
        cout << n << endl;
        return;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
void solve() {
  string n, v;
  int k;
  cin >> n >> k;
  while (true) {
    set<char> s;
    int len = n.length();
    for (int i = 0; i < len; i++) {
      s.insert(n[i]);
      if (s.size() > k) {
        while (n[i] == '9') i--;
        n[i]++;
        for (int j = i + 1; j < n.size(); j++) n[j] = '0';
        break;
      }
      if (i == len - 1) {
        cout << n << endl;
        return;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

```