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:
from sys import stdin, stdout
input = stdin.readline
print = lambda x:stdout.write(str(x)+'\n')


def findsum(d):
    sums = 0
    while d:
        sums += d%10
        d//=10
    return sums



for _ in range(int(input())):
    n, s = map(int,input().split())
    #n = list(map(int, str(n)))[::-1]
    upd = n
    i=len(str(upd))-1
    
    while findsum(upd)>s:
        t = str(upd)
        while i!=-1 and t[i]=='0':
            i-=1
        if i==-1:
            upd = 10**len(t)
            break
        upd -= int(t[i])*(10**(len(t)-i-1))
        upd += 10**(len(t)-i)
    print(upd-n)