#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
void solve() {
  int n, k;
  cin >> n >> k;
  if (n == 1) {
    cout << 1 << '\n';
    return;
  }
  n--;
  string s = to_string(n);
  string ans;
  for (int i = s.size() - 1; i >= 0; i--) {
    int dig = s[i] - '0';
    for (int j = dig + 1; j <= 9; j++) {
      set<int> gg;
      gg.insert(j);
      for (int x = 0; x < i; x++) gg.insert(s[x] - '0');
      if (gg.size() > k) continue;
      if ((int)gg.size() < k) {
        cout << s.substr(0, i) + (char)('0' + j) + string(s.size() - 1 - i, '0')
             << '\n';
        return;
      } else {
        int min_dig = 9;
        min_dig = min(min_dig, j);
        for (int p = 0; p < i; p++) min_dig = min(min_dig, s[p] - '0');
        cout << s.substr(0, i) + (char)('0' + j) +
                    string(s.size() - 1 - i, min_dig + '0')
             << '\n';
        return;
      }
    }
  }
  if (k == 1) {
    cout << string(s.size() + 1, '1') << '\n';
  } else {
    cout << '1' + string(s.size(), '0') << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int tst;
  cin >> tst;
  while (tst--) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
#pragma GCC optimize("O3")
using namespace std;
void solve() {
  int n, k;
  cin >> n >> k;
  if (n == 1) {
    cout << 1 << '\n';
    return;
  }
  n--;
  string s = to_string(n);
  string ans;
  for (int i = s.size() - 1; i >= 0; i--) {
    int dig = s[i] - '0';
    for (int j = dig + 1; j <= 9; j++) {
      set<int> gg;
      gg.insert(j);
      for (int x = 0; x < i; x++) gg.insert(s[x] - '0');
      if (gg.size() > k) continue;
      if ((int)gg.size() < k) {
        cout << s.substr(0, i) + (char)('0' + j) + string(s.size() - 1 - i, '0')
             << '\n';
        return;
      } else {
        int min_dig = 9;
        min_dig = min(min_dig, j);
        for (int p = 0; p < i; p++) min_dig = min(min_dig, s[p] - '0');
        cout << s.substr(0, i) + (char)('0' + j) +
                    string(s.size() - 1 - i, min_dig + '0')
             << '\n';
        return;
      }
    }
  }
  if (k == 1) {
    cout << string(s.size() + 1, '1') << '\n';
  } else {
    cout << '1' + string(s.size(), '0') << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int tst;
  cin >> tst;
  while (tst--) {
    solve();
  }
  return 0;
}

```