#include <bits/stdc++.h>
using namespace std;
int k, cnt[1024], len, n, ans;
char st[15];
void dfs(bool limit, int pos, int Bit, int x) {
  if (ans != INT_MAX) return;
  if (pos == len + 1) {
    if (x >= n) ans = min(ans, x);
    return;
  }
  int min = INT_MAX;
  for (int i = (limit ? (st[pos] - '0') : 0); i <= 9; ++i) {
    if (cnt[Bit | (1 << i)] <= k)
      dfs(limit & (i == (st[pos] - '0')), pos + 1, Bit | (1 << i), x * 10 + i);
  }
}
int main() {
  int tmp;
  scanf("%d", &tmp);
  cnt[0] = 0;
  for (int i = 1; i < 1024; ++i) cnt[i] = cnt[i - (i & (-i))] + 1;
  while (tmp--) {
    scanf("%s", st + 1);
    len = strlen(st + 1);
    n = 0;
    for (int i = 1; i <= len; ++i) n = n * 10 + (st[i] - '0');
    scanf("%d", &k);
    ans = INT_MAX;
    dfs(1, 1, 0, 0);
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
int k, cnt[1024], len, n, ans;
char st[15];
void dfs(bool limit, int pos, int Bit, int x) {
  if (ans != INT_MAX) return;
  if (pos == len + 1) {
    if (x >= n) ans = min(ans, x);
    return;
  }
  int min = INT_MAX;
  for (int i = (limit ? (st[pos] - '0') : 0); i <= 9; ++i) {
    if (cnt[Bit | (1 << i)] <= k)
      dfs(limit & (i == (st[pos] - '0')), pos + 1, Bit | (1 << i), x * 10 + i);
  }
}
int main() {
  int tmp;
  scanf("%d", &tmp);
  cnt[0] = 0;
  for (int i = 1; i < 1024; ++i) cnt[i] = cnt[i - (i & (-i))] + 1;
  while (tmp--) {
    scanf("%s", st + 1);
    len = strlen(st + 1);
    n = 0;
    for (int i = 1; i <= len; ++i) n = n * 10 + (st[i] - '0');
    scanf("%d", &k);
    ans = INT_MAX;
    dfs(1, 1, 0, 0);
    printf("%d\n", ans);
  }
  return 0;
}

```