Problem: You are given a positive integer n. In one move, you can increase n by one (i.e. make n := n + 1). Your task is to find the minimum number of moves you need to perform in order to make the sum of digits of n be less than or equal to s.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains two integers n and s (1 ≤ n ≤ 10^{18}; 1 ≤ s ≤ 162).

Output

For each test case, print the answer: the minimum number of moves you need to perform in order to make the sum of digits of n be less than or equal to s.

Example

Input


5
2 1
1 1
500 4
217871987498122 10
100000000000000001 1


Output


8
0
500
2128012501878
899999999999999999

Solution (in PYTHON3):

Solution:
t=int(input())

for _ in range(t):
    n,s = list(map(int,input().split()))
    n=str(n)
    arr = [int(n[i]) for i in range(len(n)-1,-1,-1)]
    count = 0
    checks = sum(arr)
    arr.append(0)
    arr.append(0)
    arr.append(0)
    arr.append(0)
    arr.append(0)
    arr.append(0)
    arr.append(0)
    arr.append(0)
    arr.append(0)
    if checks<=s:
        print(0)
    
    else:
        for i in range(0,len(n)):

            if checks<=s:
                break
            else:
                if arr[i] == 10:
                    arr[i+1] = arr[i+1]+1
                    checks = checks - arr[i] + 1
                    arr[i] = 0
                if arr[i]!=0:
                    x=10-arr[i]
                    arr[i] = 10
                    checks = checks +x
                    count+=x*(10**i) 
                if arr[i] == 10:
                    checks = checks - arr[i] + 1
                    arr[i] = 0
                    arr[i+1] = arr[i+1]+1
        
        print(count)