human
PYTHON3 solution for "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":



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Write me a Python program to find the minimum integer x such that x ≥ n and the number x is k-beautiful?
assistant
#Here is Python program to find the minimum integer x such that x ≥ n and the number x is k-beautiful.
 def solve():
    n, k = map(int, input().split())
    s = str(n)
    se = set()
    for i in range(len(s)):
        se.add(s[i])
    if len(se) <= k:
        return s
    se = set()
    i = 0
    while True:
        se.add(s[i])
        if len(se) > k:
            break
        i += 1
    se.remove(s[i])
    arr = sorted(se)
    while True:
        cur = set(c for c in s[:i])
        if len(cur) == k:
            if s[i] >= arr[-1]:
                i -= 1
                continue
            j = 0
            while arr[j] <= s[i]:
                j += 1
            return s[:i] + arr[j] + arr[0]*(len(s)-i-1)
        c = str(int(s[i])+1)
        cur.add(c)
        r = min(cur) if len(cur) == k else "0"
        return s[:i] + c + r*(len(s)-i-1)


import sys
input = lambda: sys.stdin.readline().rstrip()
t = int(input())
for i in range(t):
    print(solve())