#include <bits/stdc++.h>
using namespace std;
const int maxn = 200009;
vector<int> a;
string in;
int k;
string build(string s, vector<char> ch, int ind, bool bigger) {
  char nw = s[ind];
  if (ind == s.size()) return s;
  string tmp;
  if (bigger) {
    for (char c = '0'; c <= '9'; ++c) {
      bool check = true;
      for (int i = 0; i < ch.size() && check; ++i)
        if (ch[i] == c) check = false;
      if (check == false) {
        s[ind] = c;
        tmp = build(s, ch, ind + 1, 1);
        if (tmp != "-1") return tmp;
        s[ind] = nw;
      } else {
        if (ch.size() == k) continue;
        ch.push_back(c);
        s[ind] = c;
        tmp = build(s, ch, ind + 1, 1);
        s[ind] = nw;
        if (tmp != "-1") return tmp;
        ch.pop_back();
      }
    }
  } else {
    for (char c = nw; c <= '9'; ++c) {
      bool check = true;
      for (int i = 0; i < ch.size() && check; ++i)
        if (ch[i] == c) check = false;
      if (check == true) {
        if (ch.size() == k) continue;
        ch.push_back(c);
        s[ind] = c;
        if (c == nw)
          tmp = build(s, ch, ind + 1, 0);
        else
          tmp = build(s, ch, ind + 1, 1);
        if (tmp != "-1") return tmp;
        s[ind] = nw;
        ch.pop_back();
      } else {
        s[ind] = c;
        if (c == nw)
          tmp = build(s, ch, ind + 1, 0);
        else
          tmp = build(s, ch, ind + 1, 1);
        s[ind] = nw;
        if (tmp != "-1") return tmp;
      }
    }
  }
  return "-1";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> in >> k;
    vector<char> v;
    cout << build(in, v, 0, 0) << endl;
  }
  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>
using namespace std;
const int maxn = 200009;
vector<int> a;
string in;
int k;
string build(string s, vector<char> ch, int ind, bool bigger) {
  char nw = s[ind];
  if (ind == s.size()) return s;
  string tmp;
  if (bigger) {
    for (char c = '0'; c <= '9'; ++c) {
      bool check = true;
      for (int i = 0; i < ch.size() && check; ++i)
        if (ch[i] == c) check = false;
      if (check == false) {
        s[ind] = c;
        tmp = build(s, ch, ind + 1, 1);
        if (tmp != "-1") return tmp;
        s[ind] = nw;
      } else {
        if (ch.size() == k) continue;
        ch.push_back(c);
        s[ind] = c;
        tmp = build(s, ch, ind + 1, 1);
        s[ind] = nw;
        if (tmp != "-1") return tmp;
        ch.pop_back();
      }
    }
  } else {
    for (char c = nw; c <= '9'; ++c) {
      bool check = true;
      for (int i = 0; i < ch.size() && check; ++i)
        if (ch[i] == c) check = false;
      if (check == true) {
        if (ch.size() == k) continue;
        ch.push_back(c);
        s[ind] = c;
        if (c == nw)
          tmp = build(s, ch, ind + 1, 0);
        else
          tmp = build(s, ch, ind + 1, 1);
        if (tmp != "-1") return tmp;
        s[ind] = nw;
        ch.pop_back();
      } else {
        s[ind] = c;
        if (c == nw)
          tmp = build(s, ch, ind + 1, 0);
        else
          tmp = build(s, ch, ind + 1, 1);
        s[ind] = nw;
        if (tmp != "-1") return tmp;
      }
    }
  }
  return "-1";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> in >> k;
    vector<char> v;
    cout << build(in, v, 0, 0) << endl;
  }
  return 0;
}

```