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:
#: Author - Soumya Saurav
import sys,io,os,time
from collections import defaultdict
from collections import Counter
from collections import deque
from itertools import combinations
from itertools import permutations
import bisect,math,heapq
alphabet = "abcdefghijklmnopqrstuvwxyz"

input = sys.stdin.readline

########################################
def sod(num):
	temp = 0
	while num:
		temp += num%10
		num = num // 10
	return temp
for ii in range(int(input())):
	n , s = map(int , input().split())
	kk = n
	def f(dig, ind):
		#print("!!",dig,ind,"sum : ",sod(dig))#
		if sod(dig) <= s:
			return 0
		else:
			temp = 10 - (dig%10) if dig%10 != 0 else 0
			ans = temp * (10**ind)
			#print("yeet : ",ans)
			ans += f(dig//10 + (1 if temp > 0 else 0) , ind+1)
			return ans
	print(f(n,0))


'''
# Wrap solution for more recursion depth
#submit as python3
import collections,sys,threading
sys.setrecursionlimit(10**9)
threading.stack_size(10**8)



threading.Thread(target=solve).start()
'''