#include <bits/stdc++.h>
int64_t n, k, i;
std::string s, y;
std::set<char> st;
std::string solve() {
  st.clear();
  n = s.length();
  y = "";
  for (i = 0; i < n && st.size() <= k; ++i) {
    y += s[i];
    st.insert(s[i]);
  }
  if (i == n && st.size() <= k) return y;
  s = std::to_string(std::stoi(y) + 1) + std::string(n - y.size(), '0');
  return solve();
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  int t;
  std::cin >> t;
  while (t--) {
    std::cin >> s >> k;
    std::cout << solve() << '\n';
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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>
int64_t n, k, i;
std::string s, y;
std::set<char> st;
std::string solve() {
  st.clear();
  n = s.length();
  y = "";
  for (i = 0; i < n && st.size() <= k; ++i) {
    y += s[i];
    st.insert(s[i]);
  }
  if (i == n && st.size() <= k) return y;
  s = std::to_string(std::stoi(y) + 1) + std::string(n - y.size(), '0');
  return solve();
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  int t;
  std::cin >> t;
  while (t--) {
    std::cin >> s >> k;
    std::cout << solve() << '\n';
  }
  return 0;
}

```