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:
a=int(input())
i=1
while(i<a+1):
    b,c=input().split()
    j=0
    sum=0
    while(j<len(b)):
        sum+=int(b[j])
        j+=1
    num=[]
    for x in b:
        num+=(x)
    ans=0
    k=1

    while(sum>int(c) and len(num)-k>0):
        if(int(num[len(num)-k])>0):
            ans+=(10-int(num[len(num)-k]))*(10**(k-1))
            sum+=1-int(num[len(num)-k])
            num[len(num) - k]=0
            num[len(num)-k-1]=(int(num[len(num)-k-1])+1)

        k+=1

    if(k<=len(num) and sum<=int(c)):
         print(ans)
    else:
        if(int(b)!=int(c) and sum!=int(c)):
           print(int(10**len(num))-int(b))
        else:
            print("0")
    i+=1