#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 10;
const int maxm = 2e7 + 10;
int calc(int first) {
  int cnt = 0;
  long long res = 0;
  while (first) {
    res |= (1 << (first % 10));
    first /= 10;
  }
  while (res) {
    res -= res & -res;
    ++cnt;
  }
  return cnt;
}
int main() {
  ios::sync_with_stdio(false);
  int __;
  cin >> __;
  while (__--) {
    long long n, k;
    cin >> n >> k;
    while (calc(n) > k) {
      long long f = n, t = 1;
      while (calc(f / 10) > k) {
        f /= 10;
        t *= 10;
      };
      ++f;
      n = f * t;
    }
    cout << n << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const double pi = acos(-1.0);
const double eps = 1e-10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 10;
const int maxm = 2e7 + 10;
int calc(int first) {
  int cnt = 0;
  long long res = 0;
  while (first) {
    res |= (1 << (first % 10));
    first /= 10;
  }
  while (res) {
    res -= res & -res;
    ++cnt;
  }
  return cnt;
}
int main() {
  ios::sync_with_stdio(false);
  int __;
  cin >> __;
  while (__--) {
    long long n, k;
    cin >> n >> k;
    while (calc(n) > k) {
      long long f = n, t = 1;
      while (calc(f / 10) > k) {
        f /= 10;
        t *= 10;
      };
      ++f;
      n = f * t;
    }
    cout << n << '\n';
  }
  return 0;
}

```