#include <bits/stdc++.h>
int funcK(int v) {
  int num[10];
  for (int i = 0; i < 10; ++i) num[i] = 0;
  while (v > 0) {
    num[v % 10] = 1;
    v /= 10;
  }
  int x = 0;
  for (int i = 0; i < 10; ++i) x += num[i];
  return x;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, k;
    scanf("%d%d", &n, &k);
    while (funcK(n) > k) {
      int o = n;
      while (funcK(o) > k) o /= 10;
      int p = 1;
      while (n / p != o) p *= 10;
      p /= 10;
      n = (n / p) * p + p;
    }
    printf("%d\n", n);
  }
}

### Prompt

Generate 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>
int funcK(int v) {
  int num[10];
  for (int i = 0; i < 10; ++i) num[i] = 0;
  while (v > 0) {
    num[v % 10] = 1;
    v /= 10;
  }
  int x = 0;
  for (int i = 0; i < 10; ++i) x += num[i];
  return x;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, k;
    scanf("%d%d", &n, &k);
    while (funcK(n) > k) {
      int o = n;
      while (funcK(o) > k) o /= 10;
      int p = 1;
      while (n / p != o) p *= 10;
      p /= 10;
      n = (n / p) * p + p;
    }
    printf("%d\n", n);
  }
}

```