#include <bits/stdc++.h>
using namespace std;
long long t, k, n, ans[21], cnt[21];
string s;
bool dfs(long long now, long long val, bool tag) {
  if (val > k) return 0;
  if (now == n && val <= k) {
    for (register long long i = 0; i < n; ++i) cout << ans[i];
    cout << endl;
    return 1;
  }
  for (register long long i = tag ? s[now] - '0' : 0; i <= 9; ++i) {
    ++cnt[i];
    ans[now] = i;
    if (cnt[i] == 1) {
      if (dfs(now + 1, val + 1, tag && i == s[now] - '0')) return 1;
    } else if (dfs(now + 1, val, tag && i == s[now] - '0'))
      return 1;
    --cnt[i];
  }
  return 0;
}
signed main() {
  cin >> t;
  while (t--) {
    for (register long long i = 0; i <= 10; ++i) cnt[i] = 0;
    for (register long long i = 0; i <= 10; ++i) ans[i] = 0;
    cin >> s >> k;
    n = s.length();
    if (!dfs(0, 0, 1)) {
      cout << 10;
      for (register long long i = 2; i < k; ++i) cout << i;
      cout << endl;
    }
  }
  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;
long long t, k, n, ans[21], cnt[21];
string s;
bool dfs(long long now, long long val, bool tag) {
  if (val > k) return 0;
  if (now == n && val <= k) {
    for (register long long i = 0; i < n; ++i) cout << ans[i];
    cout << endl;
    return 1;
  }
  for (register long long i = tag ? s[now] - '0' : 0; i <= 9; ++i) {
    ++cnt[i];
    ans[now] = i;
    if (cnt[i] == 1) {
      if (dfs(now + 1, val + 1, tag && i == s[now] - '0')) return 1;
    } else if (dfs(now + 1, val, tag && i == s[now] - '0'))
      return 1;
    --cnt[i];
  }
  return 0;
}
signed main() {
  cin >> t;
  while (t--) {
    for (register long long i = 0; i <= 10; ++i) cnt[i] = 0;
    for (register long long i = 0; i <= 10; ++i) ans[i] = 0;
    cin >> s >> k;
    n = s.length();
    if (!dfs(0, 0, 1)) {
      cout << 10;
      for (register long long i = 2; i < k; ++i) cout << i;
      cout << endl;
    }
  }
  return 0;
}

```