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
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil,pi,sin,cos,acos,atan
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from bisect import insort
from collections import Counter
from collections import deque
from heapq import heappush,heappop,heapify
from itertools import permutations,combinations
mod = int(1e9)+7

 
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin. readline().rstrip()
out = lambda x : stdout.write(str(x)+"\n")


t = ip()
for _ in range(t):
    n,s = input().split(" ")
    s = int(s)
    arr = [0]
    for i in n:
        arr.append(int(i))
    ans = 0
    pos = len(arr)-1
    power = 0
    while sum(arr)> s:
        if arr[pos] == 0:
            pos -= 1
            power += 1
        else:
            val = 10-arr[pos]
            arr[pos] =  0
            ans += val*pow(10,power)
            arr[pos-1] += 1
            pos -= 1
            power += 1
    out(ans)