#include <bits/stdc++.h>
using namespace std;
string solve() {
  string n;
  int k;
  cin >> n >> k;
  while (1) {
    set<char> s;
    for (auto c : n) s.insert(c);
    if (s.size() <= k) return n;
    s.clear();
    for (int ptr = 0;; ptr++) {
      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;
      }
    }
  }
}
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    cout << solve() << '\n';
  }
  return 0;
}

### Prompt

Your task is to create 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;
string solve() {
  string n;
  int k;
  cin >> n >> k;
  while (1) {
    set<char> s;
    for (auto c : n) s.insert(c);
    if (s.size() <= k) return n;
    s.clear();
    for (int ptr = 0;; ptr++) {
      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;
      }
    }
  }
}
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    cout << solve() << '\n';
  }
  return 0;
}

```