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:
'''t=int(input())
while t>0:
	t-=1
	n,s=map(int,input().split())
	z=n
	cnt=0
	su=0
	a=[]
	while z>0:
		k=z%10
		su+=k
		a.append(k)
		z=z//10
		#print(z)
		cnt+=1
	#print(su)
	#print(cnt)
	print(a)
	if su<=s:
		print(0)
	else:
		i=cnt-1
		summ=0
		ct=0
		o=[]
		while summ<=s:
			summ+=a[i]
			o.append(a[i])
			i=i-1
			ct+=1
			
		print(summ)
		print(ct-1)
		print(o)
		while sum(o)+1>s:
			o.pop()
		print(o)
		r=0
		j=0
		for i in range(len(o)-1,-1,-1):
			r=r+o[i]*(10**j)
			j+=1
		print(r+1)



		print((r*(10**(cnt-len(o)))-n))'''

for _ in range(int(input())):
    n,s = map(int,input().split())
    ns = str(n)
    l = len(ns)
    i = 0
    summ= 0

    p = n
    a = []
    while p>0:
        a.append(p%10)
        p//=10
    a.reverse()
    summ = sum(a)
    if summ<=s:
        print('0')
        continue
    m = 0
    i = 0
    while m+a[i]<s:
        m+=a[i]
        i+=1
    # print(m,a[i])
    j = 0
    ans = 0
    while j<i:
        ans = ans*10+a[j]
        j+=1
    ans+=1
    ans = ans*(10**(l-i))
    print(ans-n)