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:
#_______________________________________________________________#
def fact(x):
	if x == 0:
		return 1
	else:
		return x * fact(x-1)
def lower_bound(li, num): 
	answer = -1
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2
		if li[middle] >= num:
			answer = middle
			end = middle - 1
		else:
			start = middle + 1
	return answer #index where x is not less than num
def upper_bound(li, num): 
	answer = -1
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2

		if li[middle] <= num:
			answer = middle
			start = middle + 1
		
		else:
			end = middle - 1
	return answer #index where x is not greater than num

def abs(x):
	return x if x >=0 else -x
def binary_search(li, val, lb, ub): 
	ans = 0
	while(lb <= ub):
		mid = (lb+ub)//2
		#print(mid, li[mid])
		if li[mid] > val:
			ub = mid-1
		elif val > li[mid]:
			lb = mid + 1
		else:
			ans = 1
			break
	return ans
def kadane(x): #maximum sum contiguous subarray
	sum_so_far = 0
	current_sum = 0
	for i in x:
		current_sum += i
		if current_sum < 0:
			current_sum = 0
		else:
			sum_so_far = mpos(sum_so_far,current_sum)
	return sum_so_far
def pref(li):
	pref_sum = [0]
	for i in li:
		pref_sum.append(pref_sum[-1] + i)
	return pref_sum
def graph(n,m):
	adj = dict()
	for i in range(1,n+1):
		adj.setdefault(i,0)
	for i in range(m):
		a,b = map(int,input().split())
		adj[a] += 1
		adj[b] += 1
	return adj

#_______________________________________________________________#
'''
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'''
import sys
import threading
from math import *
sys.setrecursionlimit(300000)
threading.stack_size(10**8)

po = int(log(10**18))+2 #for reaching exact
li = []
for i in range(po):
	li.append(3**i)
li = li[::-1]
s = sum(li)
#def main():
for _ in range(int(input())):
#for _ in range(1):
	#n = int(input())
	#n,s = map(int,input().split())
	n, s = input().split()
	s = int(s)
	#a = list(input())
	#b = list(input())
	#a = list(map(int,input().split()))
	#b = list(map(int,input().split()))
	#adj = graph(n,m)
	a = list(n)
	a = [int(x) for x in a]
	a.insert(0,0)
	def check(li,s):
		currsum = 0
		for i in range(len(li)):
			currsum += li[i]
			if currsum > s:
				return i
		return -1
	if check(a,s) == -1:
		print(0)
	else:
		while(check(a,s) != -1):
			ind = check(a,s)
			a[ind -1] += 1
			a = a[:ind] + [0]*(len(a)-ind)
		num = 0
		for i in a:
			num = num*10 + i
		print(num-int(n))














'''		
t = threading.Thread(target=main)
t.start()
t.join()
'''