#include <bits/stdc++.h>
using namespace std;
const long long oo = INT_MAX;
long long t, n, k, cnt, a[100005], b[100005], fr[100005];
long long build(long long poz) {
  long long auxcnt = cnt;
  if (cnt > k) {
    return oo;
  }
  for (long long i = poz + 1; i <= a[0]; i++) {
    b[i] = -1;
    for (long long c = 0; c <= 9; c++) {
      if (i == poz + 1 && c <= a[i]) {
        continue;
      }
      if (fr[c] || (!fr[c] && cnt < k)) {
        ++fr[c];
        b[i] = c;
        if (fr[c] == 1) {
          ++cnt;
        }
        break;
      }
    }
  }
  for (long long i = poz + 1; i <= a[0]; i++) {
    if (b[i] == -1) {
      continue;
    }
    --fr[b[i]];
  }
  cnt = auxcnt;
  long long val = 0;
  for (long long i = 1; i <= poz; i++) {
    val = val * 10 + a[i];
  }
  for (long long i = poz + 1; i <= a[0]; i++) {
    if (b[i] == -1) {
      return oo;
    }
    val = val * 10 + b[i];
  }
  return val;
}
void solve_test() {
  cin >> n >> k;
  for (long long i = 0; i <= 9; i++) {
    fr[i] = 0;
  }
  cnt = 0;
  long long aux = n;
  a[0] = 0;
  while (n) {
    a[++a[0]] = n % 10;
    b[a[0]] = n % 10;
    n /= 10;
  }
  reverse(a + 1, a + a[0] + 1);
  long long Min = build(0);
  for (long long i = 1; i <= a[0]; i++) {
    ++fr[a[i]];
    if (fr[a[i]] == 1) {
      ++cnt;
    }
    Min = min(Min, build(i));
  }
  cout << Min << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> t;
  for (long long test = 1; test <= t; test++) {
    solve_test();
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 long long oo = INT_MAX;
long long t, n, k, cnt, a[100005], b[100005], fr[100005];
long long build(long long poz) {
  long long auxcnt = cnt;
  if (cnt > k) {
    return oo;
  }
  for (long long i = poz + 1; i <= a[0]; i++) {
    b[i] = -1;
    for (long long c = 0; c <= 9; c++) {
      if (i == poz + 1 && c <= a[i]) {
        continue;
      }
      if (fr[c] || (!fr[c] && cnt < k)) {
        ++fr[c];
        b[i] = c;
        if (fr[c] == 1) {
          ++cnt;
        }
        break;
      }
    }
  }
  for (long long i = poz + 1; i <= a[0]; i++) {
    if (b[i] == -1) {
      continue;
    }
    --fr[b[i]];
  }
  cnt = auxcnt;
  long long val = 0;
  for (long long i = 1; i <= poz; i++) {
    val = val * 10 + a[i];
  }
  for (long long i = poz + 1; i <= a[0]; i++) {
    if (b[i] == -1) {
      return oo;
    }
    val = val * 10 + b[i];
  }
  return val;
}
void solve_test() {
  cin >> n >> k;
  for (long long i = 0; i <= 9; i++) {
    fr[i] = 0;
  }
  cnt = 0;
  long long aux = n;
  a[0] = 0;
  while (n) {
    a[++a[0]] = n % 10;
    b[a[0]] = n % 10;
    n /= 10;
  }
  reverse(a + 1, a + a[0] + 1);
  long long Min = build(0);
  for (long long i = 1; i <= a[0]; i++) {
    ++fr[a[i]];
    if (fr[a[i]] == 1) {
      ++cnt;
    }
    Min = min(Min, build(i));
  }
  cout << Min << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> t;
  for (long long test = 1; test <= t; test++) {
    solve_test();
  }
  return 0;
}

```