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:
import math
def dtb(n): 
    return bin(n).replace("0b","")
def btd(n): 
    return int(n,2)
    
def sumofdigits(n):
    sod=0
    while(n>0):
        sod+=n%10
        n//=10
    return sod    
t=int(input())
for k in range(t):
    n,s=map(int,input().split())
    m=n
    sod=sumofdigits(m)
    if sod<=s:
        print(0)
        continue
    fact=10
    while(1):
        rem=n%fact
        n+=(fact-rem)%fact
        sod=sumofdigits(n)
        if(sod<=s):
            break
        fact*=10
    print(n-m)