#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
string get(long long i, long long n, long long dis, long long k, string org,
           string cur) {
  if (dis > k) {
    return "";
  }
  if (cur.substr(0, i) < org.substr(0, i)) {
    return "";
  }
  if (i >= n) {
    return cur;
  }
  for (char x = '0'; x <= '9'; x += 1) {
    cur[i] = x;
    bool check = false;
    for (long long j = 0; j < i; j++) {
      if (cur[j] == x) {
        check = true;
        break;
      }
    }
    string lol = get(i + 1, n, dis + (!check), k, org, cur);
    if (lol.size()) {
      return lol;
    }
  }
  return "";
}
void solve() {
  long long n, k;
  cin >> n >> k;
  string s = to_string(n);
  string ans = get(0, s.size(), 0, k, s, string(s.size(), '0'));
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
string get(long long i, long long n, long long dis, long long k, string org,
           string cur) {
  if (dis > k) {
    return "";
  }
  if (cur.substr(0, i) < org.substr(0, i)) {
    return "";
  }
  if (i >= n) {
    return cur;
  }
  for (char x = '0'; x <= '9'; x += 1) {
    cur[i] = x;
    bool check = false;
    for (long long j = 0; j < i; j++) {
      if (cur[j] == x) {
        check = true;
        break;
      }
    }
    string lol = get(i + 1, n, dis + (!check), k, org, cur);
    if (lol.size()) {
      return lol;
    }
  }
  return "";
}
void solve() {
  long long n, k;
  cin >> n >> k;
  string s = to_string(n);
  string ans = get(0, s.size(), 0, k, s, string(s.size(), '0'));
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```