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":
assistant
#Here is Python code to find the minimum integer x such that x ≥ n and the number x is k-beautiful.
 def upList(s, i):
    left = s[:i]
    right = s[i:]
    left = list(str(int(''.join(left)) + 1))
    return left + right

def jg(x, val):
    y = set(i for i in x if i >= val)
    if len(y) == 0:
        return None
    return min(y)

for _ in range(int(input())):
    n, k = map(int, input().split())
    s = list(str(n))

    i = len(s) - 1
    while i >= 0:
        cMax = set()
        j = 0
        while len(cMax) < k and j < len(s):
            cMax.add(s[j])
            j += 1
        if j == len(s):
            break

        if s[i] not in cMax:
            z = jg(cMax, s[i])
            if z == None:
                s = upList(s, i)
                for var in range(i, len(s)):
                    s[var] = '0'
                i = len(s) - 1
            else:
                s[i] = z
                for var in range(i + 1, len(s)):
                    s[var] = min(cMax)
                i -= 1
        else:
            i -= 1
        
    print(''.join(s))