#include <bits/stdc++.h>
using namespace std;
long long digits(long long x) {
  long long y = 0;
  while (x) {
    y |= 1 << (x % 10);
    x /= 10;
  }
  return __builtin_popcount(y);
}
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
  }
  long long i, j, k, n, m, c, t, d;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    while (digits(n) > k) {
      c = n / 10;
      d = 10;
      while (digits(c) > k) {
        c /= 10;
        d *= 10;
      }
      d /= 10;
      n = ((n / d) + 1) * d;
    }
    cout << n << "\n";
  }
}

### Prompt

In Cpp, your task is to solve 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 digits(long long x) {
  long long y = 0;
  while (x) {
    y |= 1 << (x % 10);
    x /= 10;
  }
  return __builtin_popcount(y);
}
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
  }
  long long i, j, k, n, m, c, t, d;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    while (digits(n) > k) {
      c = n / 10;
      d = 10;
      while (digits(c) > k) {
        c /= 10;
        d *= 10;
      }
      d /= 10;
      n = ((n / d) + 1) * d;
    }
    cout << n << "\n";
  }
}

```