#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int t, n, k, st[30], vis[32];
int main() {
  scanf("%lld", &t);
  while (t--) {
    scanf("%d%d", &n, &k);
    st[0] = 0;
    while (n) st[++st[0]] = n % 10, n /= 10;
    memset(vis, 0, sizeof(vis));
    while (1) {
      int status = 0, num = 0;
      for (register int i = 1; i <= st[0]; ++i)
        if (st[i] >= 10) st[i] = 0, st[i + 1]++;
      for (register int i = st[0]; i >= 1; --i) {
        num += !(status >> st[i] & 1);
        status |= 1 << st[i];
        if (num > k) {
          ++st[i];
          for (register int j = 1; j <= i - 1; ++j) st[j] = 0;
          break;
        }
      }
      if (num <= k) {
        for (register int i = st[0]; i >= 1; --i) printf("%d", st[i]);
        puts("");
        break;
      }
    }
  }
  return 0;
}

### 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;
const int N = 100005;
int t, n, k, st[30], vis[32];
int main() {
  scanf("%lld", &t);
  while (t--) {
    scanf("%d%d", &n, &k);
    st[0] = 0;
    while (n) st[++st[0]] = n % 10, n /= 10;
    memset(vis, 0, sizeof(vis));
    while (1) {
      int status = 0, num = 0;
      for (register int i = 1; i <= st[0]; ++i)
        if (st[i] >= 10) st[i] = 0, st[i + 1]++;
      for (register int i = st[0]; i >= 1; --i) {
        num += !(status >> st[i] & 1);
        status |= 1 << st[i];
        if (num > k) {
          ++st[i];
          for (register int j = 1; j <= i - 1; ++j) st[j] = 0;
          break;
        }
      }
      if (num <= k) {
        for (register int i = st[0]; i >= 1; --i) printf("%d", st[i]);
        puts("");
        break;
      }
    }
  }
  return 0;
}

```