#include <bits/stdc++.h>
using namespace std;
char s[12];
int len, ans, k;
int cal(int mask) { return __builtin_popcount(mask); }
void dfs(int lim, int pos, int mask, int sum) {
  if (pos == len + 1) {
    ans = sum;
    return;
  }
  int start = 0;
  if (lim) {
    start = s[pos];
  }
  for (int i = start; i <= 9; i++) {
    if (ans == 0 && cal(mask | (1 << i)) <= k) {
      dfs(lim && i == start, pos + 1, mask | (1 << i), sum * 10 + i);
    }
  }
}
int main() {
  int t;
  int n;
  cin >> t;
  while (t--) {
    scanf("%s", s + 1);
    cin >> k;
    len = strlen(s + 1);
    for (int i = 1; i <= len; i++) {
      s[i] -= '0';
    }
    ans = 0;
    dfs(1, 1, 0, 0);
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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;
char s[12];
int len, ans, k;
int cal(int mask) { return __builtin_popcount(mask); }
void dfs(int lim, int pos, int mask, int sum) {
  if (pos == len + 1) {
    ans = sum;
    return;
  }
  int start = 0;
  if (lim) {
    start = s[pos];
  }
  for (int i = start; i <= 9; i++) {
    if (ans == 0 && cal(mask | (1 << i)) <= k) {
      dfs(lim && i == start, pos + 1, mask | (1 << i), sum * 10 + i);
    }
  }
}
int main() {
  int t;
  int n;
  cin >> t;
  while (t--) {
    scanf("%s", s + 1);
    cin >> k;
    len = strlen(s + 1);
    for (int i = 1; i <= len; i++) {
      s[i] -= '0';
    }
    ans = 0;
    dfs(1, 1, 0, 0);
    cout << ans << endl;
  }
  return 0;
}

```