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:
import math
from sys import stdin
g   = lambda : stdin.readline().strip()
gl  = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")

t, = gil()

pow = []
for i in range(19):
	pow.append(10**i)


def getSum(n):
	p = math.ceil(math.log10(n)) ; ans = 0
	for p in range(p+1):
		if n:
			dig = (n // pow[p])%10
			ans += dig 
			n -= dig*(pow[p])
		else:
			break
	return ans\



for _  in range(t):
	n, s = gil()
	old = n
	cs = sum([int(var) for var in str(n)] )

	maxPow = math.ceil(math.log10(n))
	for p in range(maxPow+1):
		if getSum(n) <= s : break
		dig = (n//pow[p])%10
		toAdd = (10-dig)%10
		n += toAdd*pow[p]

	print(n - old)