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 sys
input=sys.stdin.readline
from math import ceil
def sumdig(n):
	n=str(n)
	ans=0
	for i in n:
		ans+=int(i)
	return ans
t=int(input())
while(t!=0):
	t-=1
	n,s=map(int,input().split())
	ans=0
	# if(sumdig(n)>s):
	# 	print(10**(len(str(n)))-n)
	# 	continue
	if (sumdig(n)<=s):
		print(0)
		continue

	i=1
	ans=0
	nn=n
	while (sumdig(nn)>s):
		nn=(10**i)*(nn//(10**i))+10**(i)
		i+=1


	print(nn-n)
	# sn=str(n)
	# i=len(sn)-1
	# count=0
	# while(i>0 and su>s):
	# 	su-=(int(sn[i]))
	# 	i-=1
	# 	count+=1

	# num=sn[:i]+str(int(sn[i])+1)+"0"*(count)
	# print(int(num)-n)
	# while(i<len(n) and s<su+int(n[i])):
	# 	su=su+int(n[i])
	# 	i+=1
	# if i==len(n):
	# 	num=int("1"+"0"*(i))
	# 	ans=int(num)-int(n)
	# 	print(ans)
	# else:
	# 	num=int(n[0:i]+"0"*(len(n)-i+1))
	# 	ans=int(num)-int(n)
	# 	print(ans)
	# while(sumdig(n)>s):
	# 	i=ceil((sumdig(n)-s)//10)
	# 	sn=(10**i)*(n//(10**i)+(10**(i-1)))
	# 	ans+=(sn-n)
	# 	n=sn
	# print(ans)