#include <bits/stdc++.h>
using namespace std;
const int S = int(2e5) + 100;
const int INF = int(1e9) + 7;
const long long MOD = long long(1e9) + 7;
const double PI = 3.141592653589793238;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
void print(T& v) {
  for (auto& x : v) cout << x << " ";
  cout << endl;
}
long long ans(long long n, int k) {
  vector<int> dg;
  long long on = n;
  while (n) {
    dg.emplace_back(n % 10);
    n /= 10;
  }
  reverse(dg.begin(), dg.end());
  int i = 0, seen[10] = {}, cnt = 0;
  for (; i < (int)(dg).size(); i++) {
    if (!seen[dg[i]]) {
      cnt++;
      seen[dg[i]] = 1;
    }
    if (cnt > k) break;
  }
  if (cnt <= k) return on;
  long long m = 0;
  for (int j = 0; j <= i; j++) m = m * 10 + dg[j];
  m++;
  for (int j = (0); j < ((int)(dg).size() - i - 1); ++j) m *= 10;
  return ans(m, k);
}
int t, k;
long long n;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    cout << ans(n, k) << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 S = int(2e5) + 100;
const int INF = int(1e9) + 7;
const long long MOD = long long(1e9) + 7;
const double PI = 3.141592653589793238;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
void print(T& v) {
  for (auto& x : v) cout << x << " ";
  cout << endl;
}
long long ans(long long n, int k) {
  vector<int> dg;
  long long on = n;
  while (n) {
    dg.emplace_back(n % 10);
    n /= 10;
  }
  reverse(dg.begin(), dg.end());
  int i = 0, seen[10] = {}, cnt = 0;
  for (; i < (int)(dg).size(); i++) {
    if (!seen[dg[i]]) {
      cnt++;
      seen[dg[i]] = 1;
    }
    if (cnt > k) break;
  }
  if (cnt <= k) return on;
  long long m = 0;
  for (int j = 0; j <= i; j++) m = m * 10 + dg[j];
  m++;
  for (int j = (0); j < ((int)(dg).size() - i - 1); ++j) m *= 10;
  return ans(m, k);
}
int t, k;
long long n;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    cout << ans(n, k) << endl;
  }
  return 0;
}

```