#include <bits/stdc++.h>
using namespace std;
bool run(int i, string &s, string &ans, int k, unordered_map<int, int> &m,
         bool greater) {
  int n = s.length();
  if (i == n) return true;
  int d = greater ? 0 : s[i] - '0';
  while (d < 10) {
    if (m.find(d) != m.end() || m.size() < k) {
      m[d]++;
      ans[i] = d + '0';
      bool f = run(i + 1, s, ans, k, m, greater || (s[i] - '0' < d));
      if (f) return true;
      m[d]--;
      if (m[d] == 0) m.erase(d);
    }
    d++;
  }
  return false;
}
int solve(int n, int k) {
  unordered_map<int, int> m;
  int x = log10(n);
  x = pow(10, x);
  string s = to_string(n);
  string ans = to_string(x);
  run(0, s, ans, k, m, false);
  return stoi(ans);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    cout << solve(n, k) << endl;
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
bool run(int i, string &s, string &ans, int k, unordered_map<int, int> &m,
         bool greater) {
  int n = s.length();
  if (i == n) return true;
  int d = greater ? 0 : s[i] - '0';
  while (d < 10) {
    if (m.find(d) != m.end() || m.size() < k) {
      m[d]++;
      ans[i] = d + '0';
      bool f = run(i + 1, s, ans, k, m, greater || (s[i] - '0' < d));
      if (f) return true;
      m[d]--;
      if (m[d] == 0) m.erase(d);
    }
    d++;
  }
  return false;
}
int solve(int n, int k) {
  unordered_map<int, int> m;
  int x = log10(n);
  x = pow(10, x);
  string s = to_string(n);
  string ans = to_string(x);
  run(0, s, ans, k, m, false);
  return stoi(ans);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    cout << solve(n, k) << endl;
  }
  return 0;
}

```