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 sys
input=sys.stdin.readline
#t=int(input())
import math
for _ in range(int(input())):
    n,s=map(int,input().split())
    l=list(map(int,list(str(n))))
    #print(l)
    x=0
    i=0
    while(i<len(l) and x<s):
        if x+l[i]<s:
            x+=l[i]
            i+=1
        else:
            break
    #print(x)
    if sum(l)==s:
        print(0)
    elif i==len(l):
        print(0)
    elif i==0:
        print(pow(10,len(l))-n)
    else:
        #l=l[i:]
        #print(l)
        '''
        c=0
        f=0
        for j in range(len(l)):
            if j==i-1:
                if l[j]<9:
                    l[j]=l[j]+1
                else:
                    f=1
                    l[j]=0
            elif j>=i:
                l[j]=0
            l[j]=str(l[j])
        if f:
            l.insert(0,'1')
        #print(l)
        print(int(''.join(l))-n)
        #print(''.join(l))
        '''
        for j in range(len(l)):
            l[j]=str(l[j])
        k=int(''.join(l[:i]))
        k+=1
        print(k*pow(10,len(l)-i)-n)