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 digitsum(x):
    res=0
    while(x>0):
        res+=x%10
        x//=10
    return res

def solve():
    res=10**30
    n,s=map(int,input().split())
    if digitsum(n)<=s:
        return 0
    N=len(str(n))
    for i in range(1,N+1):
        P=pow10[i]-(n%pow10[i])
        if digitsum(n+P)<=s:
            res=min(res,P)
    return res
    

t=int(input())
pow10=[1]
for i in range(20):
    pow10.append(pow10[-1]*10)
for i in range(t):
    print(solve())