#include <bits/stdc++.h>
using namespace std;
int n, k;
int pos;
vector<int> num;
void init() {
  int _n = n;
  while (_n > 0) {
    num.push_back(_n % 10);
    _n /= 10;
  }
  reverse(num.begin(), num.end());
}
long long ans;
int main() {
  int t;
  cin >> t;
  while (t--) {
    num.clear();
    cin >> n >> k;
    init();
    int msk = 0;
    for (int i = 0; i < (int)num.size(); i++) msk |= (1 << num[i]);
    if (__builtin_popcount(msk) <= k) {
      printf("%d\n", n);
      continue;
    }
    ans = 1e10;
    n = (int)num.size();
    for (int i = 0; i < n; i++) {
      bool flag = 0;
      int now = 0;
      for (int j = 0; j < i; j++) now |= (1 << num[j]);
      if (__builtin_popcount(now) > k) break;
      int lft = k - __builtin_popcount(now);
      long long nans = 0;
      for (int j = 0; j < i; j++) nans = nans * 10 + num[j];
      for (int j = num[i] + 1; j <= 9; j++)
        if (now & (1 << j) || lft) {
          nans = nans * 10 + j;
          if ((now & (1 << j)) == 0) {
            now |= (1 << j);
            lft--;
          }
          flag = 1;
          break;
        }
      if (!flag) continue;
      if (!lft && (now & 1) == 0) {
        for (int j = 1; j <= 9; j++)
          if (now & (1 << j)) {
            for (int l = i + 1; l < n; l++) nans = nans * 10 + j;
            break;
          }
      } else {
        for (int l = i + 1; l < n; l++) nans = nans * 10;
      }
      ans = min(ans, nans);
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Please formulate 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;
int n, k;
int pos;
vector<int> num;
void init() {
  int _n = n;
  while (_n > 0) {
    num.push_back(_n % 10);
    _n /= 10;
  }
  reverse(num.begin(), num.end());
}
long long ans;
int main() {
  int t;
  cin >> t;
  while (t--) {
    num.clear();
    cin >> n >> k;
    init();
    int msk = 0;
    for (int i = 0; i < (int)num.size(); i++) msk |= (1 << num[i]);
    if (__builtin_popcount(msk) <= k) {
      printf("%d\n", n);
      continue;
    }
    ans = 1e10;
    n = (int)num.size();
    for (int i = 0; i < n; i++) {
      bool flag = 0;
      int now = 0;
      for (int j = 0; j < i; j++) now |= (1 << num[j]);
      if (__builtin_popcount(now) > k) break;
      int lft = k - __builtin_popcount(now);
      long long nans = 0;
      for (int j = 0; j < i; j++) nans = nans * 10 + num[j];
      for (int j = num[i] + 1; j <= 9; j++)
        if (now & (1 << j) || lft) {
          nans = nans * 10 + j;
          if ((now & (1 << j)) == 0) {
            now |= (1 << j);
            lft--;
          }
          flag = 1;
          break;
        }
      if (!flag) continue;
      if (!lft && (now & 1) == 0) {
        for (int j = 1; j <= 9; j++)
          if (now & (1 << j)) {
            for (int l = i + 1; l < n; l++) nans = nans * 10 + j;
            break;
          }
      } else {
        for (int l = i + 1; l < n; l++) nans = nans * 10;
      }
      ans = min(ans, nans);
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```