#include <bits/stdc++.h>
using namespace std;
void write(vector<int> &v) {
  for (auto i : v) cout << i << " ";
  cout << "\n";
}
void read(vector<int> &v) {
  for (auto &i : v) cin >> i;
}
const int INF = 1e9;
const int64_t INFF = 1e18;
const int N = 1e6 + 69;
int k;
string s;
string calc(int index, string ans) {
  set<char> st((ans).begin(), (ans).end());
  if ((int)(st).size() > k) return "";
  if (ans.substr(0, index) < s.substr(0, index)) return "";
  if (index >= s.length()) {
    return ans;
  }
  for (char c = '0'; c <= '9'; c++) {
    ans.push_back(c);
    string pos = calc(index + 1, ans);
    ans.pop_back();
    if (pos.length() == s.length()) return pos;
  }
}
void solve() {
  cin >> s;
  cin >> k;
  cout << calc(0, "") << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
void write(vector<int> &v) {
  for (auto i : v) cout << i << " ";
  cout << "\n";
}
void read(vector<int> &v) {
  for (auto &i : v) cin >> i;
}
const int INF = 1e9;
const int64_t INFF = 1e18;
const int N = 1e6 + 69;
int k;
string s;
string calc(int index, string ans) {
  set<char> st((ans).begin(), (ans).end());
  if ((int)(st).size() > k) return "";
  if (ans.substr(0, index) < s.substr(0, index)) return "";
  if (index >= s.length()) {
    return ans;
  }
  for (char c = '0'; c <= '9'; c++) {
    ans.push_back(c);
    string pos = calc(index + 1, ans);
    ans.pop_back();
    if (pos.length() == s.length()) return pos;
  }
}
void solve() {
  cin >> s;
  cin >> k;
  cout << calc(0, "") << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```