human
What is a Python program to find the minimum integer x such that x ≥ n and the number x is k-beautiful given the following:
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 program to find the minimum integer x such that x ≥ n and the number x is k-beautiful.
 def main(t):
 
 
    n,k = map(int, input().split())
    s = str(n)
    m = len(s)
 
 
    dic = {}
 
    ans = ""
 
    for i in range(m):
 
        if s[i] in dic:
            dic[s[i]] += 1
        else:
            dic[s[i]] = 1
      
 
 
 
        if len(dic)==k and i<m-1:
            if s[i+1:] >  max(dic)*(m-i-1):
                dic[s[i]] -= 1
                if dic[s[i]]==0:  del dic[s[i]] 
                if str(int(s[i])+1) in dic and len(dic)<k:
                    ans += str(int(s[i])+1)
                    ans += '0'*(m-i-1)
                else:
                    dic[str(int(s[i])+1)] = 1
                    ans += str(int(s[i])+1)
                    ans += min(dic)*(m-i-1)
                break
 
            else:
                ans += s[i]
                broken = i+1
                flag = True
                for j in range(i+1,m):
                    if s[j] < max(dic): broken = j
                    if s[j] not in dic:  
                        flag = False
                        break
                if flag:  broken=m 
                ans += s[i+1:broken]
                if broken==m:  break
 
                for d in range(int(s[broken])+1,10):
                    if str(d) in dic: break
                dic[str(d)] = 1
                ans += str(d)
                ans += min(dic)*(m-len(ans))
                break
 
                  
 
        ans += s[i] 
 
    print(ans)

T = int(input())
t = 1
while t<=T:
    main(t)
    t += 1