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:
for t in range(int(input())):
    # n,s = map(int,input().split())
    n,s = input().split()
    s = int(s)

    nn =['0','0','0'] + list(n)

    su = 0
    for i in n:
        su += int(i)

    ptr = 1
    # cost = 0
    while su>s:
        su-=int(nn[-ptr])
        su+=1
        nn[-ptr]='0'
        nn[-ptr-1]=str(1+int(nn[-ptr-1]))

        itr = 1
        while int(nn[-ptr-itr]) >=10:
            su-=10
            su+=int(nn[-ptr-itr])//10
            nn[-ptr-itr-1]=str(int(nn[-ptr-itr-1])+int(nn[-ptr-itr])//10)
            nn[-ptr-itr]=str(int(nn[-ptr-1])%10)
            itr+=1
        # if int(nn[-ptr-1])>=10:
            
            
            
            
        ptr+=1

    print(int(''.join(nn))-int(n))

    # nn = n
    # ctr = 0
    # cost = []
    # su = 0
    # while nn:
    #     su += nn%10
    #     cost.append((10-nn%10)*10**ctr)
    #     nn//=10
    #     ctr+=1

    # ptr = 1
    # while su>s:
    #     su-=n%10**ptr
    #     su+=1
    #     n