#include <bits/stdc++.h>
using namespace std;
const int INF = (int)(1e9) + 10;
int n, k;
int ans;
string s;
void update(int pos, int mask, int num) {
  if (pos == (int)s.size()) {
    ans = num;
    return;
  }
  if (__builtin_popcount(mask) < k) {
    for (int i = pos; i < (int)s.size(); i++) num *= 10;
    ans = num;
    return;
  }
  if (__builtin_popcount(mask) == k) {
    for (int d = 0; d < 10; d++) {
      if (((1 << d) | mask) == mask) {
        for (int i = pos; i < (int)s.size(); i++) num = num * 10 + d;
        ans = num;
        return;
      };
    }
  };
  return;
}
bool go(int pos, bool upper, int mask, int num) {
  if ((pos == (int)s.size()) || upper) {
    update(pos, mask, num);
    return true;
  }
  int low_d = (upper) ? 0 : s[pos] - '0';
  for (int d = low_d; d < 10; d++) {
    int new_mask = mask | (1 << d);
    if (__builtin_popcount(new_mask) <= k) {
      if (go(pos + 1, upper || (d > s[pos] - '0'), new_mask, num * 10 + d))
        return true;
    }
  }
  return false;
};
void solve() {
  cin >> n >> k;
  s = to_string(n);
  ans = 0;
  go(0, 0, 0, 0);
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int T = 1;
  cin >> T;
  while (T--) solve();
  return 0;
}

### Prompt

Your task is to create a Cpp solution to 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;
const int INF = (int)(1e9) + 10;
int n, k;
int ans;
string s;
void update(int pos, int mask, int num) {
  if (pos == (int)s.size()) {
    ans = num;
    return;
  }
  if (__builtin_popcount(mask) < k) {
    for (int i = pos; i < (int)s.size(); i++) num *= 10;
    ans = num;
    return;
  }
  if (__builtin_popcount(mask) == k) {
    for (int d = 0; d < 10; d++) {
      if (((1 << d) | mask) == mask) {
        for (int i = pos; i < (int)s.size(); i++) num = num * 10 + d;
        ans = num;
        return;
      };
    }
  };
  return;
}
bool go(int pos, bool upper, int mask, int num) {
  if ((pos == (int)s.size()) || upper) {
    update(pos, mask, num);
    return true;
  }
  int low_d = (upper) ? 0 : s[pos] - '0';
  for (int d = low_d; d < 10; d++) {
    int new_mask = mask | (1 << d);
    if (__builtin_popcount(new_mask) <= k) {
      if (go(pos + 1, upper || (d > s[pos] - '0'), new_mask, num * 10 + d))
        return true;
    }
  }
  return false;
};
void solve() {
  cin >> n >> k;
  s = to_string(n);
  ans = 0;
  go(0, 0, 0, 0);
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int T = 1;
  cin >> T;
  while (T--) solve();
  return 0;
}

```