#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t, k;
  string n;
  cin >> t;
  vector<int> digits(10);
  while (t-- > 0) {
    cin >> n >> k;
    std::fill(digits.begin(), digits.end(), 0);
    int cnt = 0;
    int up_idx = 1000;
    int i = 0;
    while (i < n.length()) {
      if (i > up_idx) {
        int j = 0;
        while (j < 10) {
          if (digits[j] > 0 || cnt < k) {
            n[i] = (char)(j + '0');
            if (digits[j]++ == 0) cnt++;
            break;
          }
          j++;
        }
      } else {
        int d = n[i] - '0';
        if (digits[d]++ == 0) cnt++;
        if (cnt > k) {
          int j = d + 1;
          digits[d]--;
          while (j < 10) {
            if (digits[j] > 0) {
              n[i] = (char)(j + '0');
              digits[j]++;
              cnt--;
              up_idx = i;
              break;
            }
            j++;
          }
          if (cnt > k) {
            if (--digits[n[i - 1] - '0'] == 0) cnt--;
            cnt--;
            n[i - 1]++;
            up_idx = i - 1;
            i--;
            continue;
          }
        }
      }
      i++;
    }
    cout << n << "\n";
  }
  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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t, k;
  string n;
  cin >> t;
  vector<int> digits(10);
  while (t-- > 0) {
    cin >> n >> k;
    std::fill(digits.begin(), digits.end(), 0);
    int cnt = 0;
    int up_idx = 1000;
    int i = 0;
    while (i < n.length()) {
      if (i > up_idx) {
        int j = 0;
        while (j < 10) {
          if (digits[j] > 0 || cnt < k) {
            n[i] = (char)(j + '0');
            if (digits[j]++ == 0) cnt++;
            break;
          }
          j++;
        }
      } else {
        int d = n[i] - '0';
        if (digits[d]++ == 0) cnt++;
        if (cnt > k) {
          int j = d + 1;
          digits[d]--;
          while (j < 10) {
            if (digits[j] > 0) {
              n[i] = (char)(j + '0');
              digits[j]++;
              cnt--;
              up_idx = i;
              break;
            }
            j++;
          }
          if (cnt > k) {
            if (--digits[n[i - 1] - '0'] == 0) cnt--;
            cnt--;
            n[i - 1]++;
            up_idx = i - 1;
            i--;
            continue;
          }
        }
      }
      i++;
    }
    cout << n << "\n";
  }
  return 0;
}

```