human
What is a Python code to find the minimum integer x such that x ≥ n and the number x is k-beautiful given the following:
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
Using python3 can you solve the prior task?
assistant
#Here is Python code to find the minimum integer x such that x ≥ n and the number x is k-beautiful.

import sys
input = sys.stdin.readline
from functools import lru_cache

for t in range(int(input())):
    n, k = map(int, input().split())
    s = str(n)
    m = len(s)
    p = [0] * 12
    p[0] = 1
    for i in range(1, 12): p[i] = 10 * p[i - 1]

    @lru_cache(None)
    def dp(i, tight, mask):
        if i == m: return 0
        ans = float("inf")
        for j in range(10):
            mask2 = mask | 1 << j
            if bin(mask2).count('1') > k: continue
            if tight and j < int(s[i]): continue
            tight2 = tight and j == int(s[i])
            cur = p[m - 1 - i] * j + dp(i + 1, tight2, mask2)
            if cur < ans:
                ans = cur
                break
        return ans

    ans = dp(0, 1, 0)
    print(ans)