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:
def sumOfDigit(n): 
    list_of_number = list(map(int, str(n).strip()))
    return sum(list_of_number)
for _ in range(int(input())):
    n,s = map(int,input().split())
    temp = n
    modulo = 10
    while sumOfDigit(temp)>s:
        temp = (n//modulo)*(modulo)
        temp += modulo
        modulo *= 10
    print(temp - n)