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:
"""
https://codeforces.com/contest/1409/problem/D
"""

def digit_sum(x):
	'''Return the sum of all the digits of a number'''
	ref = x
	total = 0

	while(x > 0):
		total += x % 10
		x = x // 10

	# print ('digit_sum of ', ref, ' = ', total)
	return total

count = int(input())
# print (count)

for i in range(count):
	jump = 0
	a, b = map(int, input().split())
	copy = a

	while digit_sum(copy + 1) > b:
		# print (copy)
		copy //= 10
		jump += 1

	print ( (10 ** jump) - ( a % (10 ** jump)) if digit_sum(a) > b else 0 )