You have n gifts and you want to give all of them to children. Of course, you don't want to offend anyone, so all gifts should be equal between each other. The i-th gift consists of a_i candies and b_i oranges.

During one move, you can choose some gift 1 ≤ i ≤ n and do one of the following operations:

  * eat exactly one candy from this gift (decrease a_i by one); 
  * eat exactly one orange from this gift (decrease b_i by one); 
  * eat exactly one candy and exactly one orange from this gift (decrease both a_i and b_i by one). 



Of course, you can not eat a candy or orange if it's not present in the gift (so neither a_i nor b_i can become less than zero).

As said above, all gifts should be equal. This means that after some sequence of moves the following two conditions should be satisfied: a_1 = a_2 = ... = a_n and b_1 = b_2 = ... = b_n (and a_i equals b_i is not necessary).

Your task is to find the minimum number of moves required to equalize all the given gifts.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 50) — the number of gifts. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the number of candies in the i-th gift. The third line of the test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9), where b_i is the number of oranges in the i-th gift.

Output

For each test case, print one integer: the minimum number of moves required to equalize all the given gifts.

Example

Input


5
3
3 5 6
3 2 3
5
1 2 3 4 5
5 4 3 2 1
3
1 1 1
2 2 2
6
1 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1
3
10 12 8
7 5 4


Output


6
16
0
4999999995
7

Note

In the first test case of the example, we can perform the following sequence of moves:

  * choose the first gift and eat one orange from it, so a = [3, 5, 6] and b = [2, 2, 3]; 
  * choose the second gift and eat one candy from it, so a = [3, 4, 6] and b = [2, 2, 3]; 
  * choose the second gift and eat one candy from it, so a = [3, 3, 6] and b = [2, 2, 3]; 
  * choose the third gift and eat one candy and one orange from it, so a = [3, 3, 5] and b = [2, 2, 2]; 
  * choose the third gift and eat one candy from it, so a = [3, 3, 4] and b = [2, 2, 2]; 
  * choose the third gift and eat one candy from it, so a = [3, 3, 3] and b = [2, 2, 2]. 
#_______________________________________________________________#
def fact(x):
	if x == 0:
		return 1
	else:
		return x * fact(x-1)
def lower_bound(li, num): #return 0 if all are greater or equal to
	answer = len(li)-1
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2
		if li[middle] >= num:
			answer = middle
			end = middle - 1
		else:
			start = middle + 1
	return answer #index where x is not less than num
def upper_bound(li, num): #return n-1 if all are small or equal
	answer = -1
	start = 0
	end = len(li)-1

	while(start <= end):
		middle = (end+start)//2

		if li[middle] <= num:
			answer = middle
			start = middle + 1
		
		else:
			end = middle - 1
	return answer #index where x is not greater than num

def abs(x):
	return x if x >=0 else -x
def binary_search(li, val, lb, ub): # ITS A BINARY
	ans = 0
	while(lb <= ub):
		mid = (lb+ub)//2
		#print(mid, li[mid])
		if li[mid] > val:
			ub = mid-1
		elif val > li[mid]:
			lb = mid + 1
		else:
			ans = 1
			break
	return ans

def sieve_of_eratosthenes(n):
	ans = []
	arr = [1]*(n+1)
	arr[0],arr[1], i = 0, 0, 2
	while(i*i <= n):
		if arr[i] == 1:
			j = i+i
			while(j <= n):
				arr[j] = 0
				j += i
		i += 1
	for k in range(n):
		if arr[k] == 1:
			ans.append(k)
	return ans
def nc2(x):
	if x == 1:
		return 0
	else:
		return x*(x-1)//2
def kadane(x): #maximum subarray sum
	sum_so_far = 0
	current_sum = 0
	for i in x:
		current_sum += i
		if current_sum < 0:
			current_sum = 0
		else:
			sum_so_far = max(sum_so_far,current_sum)
	return sum_so_far
def mex(li):
	check = [0]*1001
	for i in li:
		check[i] += 1
	for i in range(len(check)):
		if check[i] == 0:
		 return i 
def sumdigits(n):
	ans = 0
	while(n!=0):
		ans += n%10
		n //= 10
	return ans

def product(li):
	ans = 1
	for i in li:
		ans *= i
	return ans
def maxpower(n,k):
	cnt = 0
	while(n>1):
		cnt += 1
		n //= k
	return cnt
def knapsack(li,sumi,s,n,dp):
	if s == sumi//2:
		return 1
	elif n==0:
		return 0
	else:
		if dp[n][s] != -1:
			return dp[n][s]
		else:
			if li[n-1] <= s:
				dp[n][s] = knapsack(li,sumi,s-li[n-1],n-1,dp) or knapsack(li,sumi,s,n-1,dp)
			else:
				dp[n][s] = knapsack(li,sumi,s,n-1,dp)
			return dp[n][s]
def pref(li):
	pref_sum = [0]
	for i in li:
		pref_sum.append(pref_sum[-1] + i)
	return pref_sum

#_______________________________________________________________#
'''
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'''
from math import *
from collections import deque
max_steps = 100005
for _ in range(int(input())):
#for _ in range(1):
	n = int(input())
	#n,k,z = map(int,input().split())
	#s1 = input()
	#s2 = input()
	#n,k = map(int,input().split())
	a = list(map(int,input().split()))
	b = list(map(int,input().split()))
	#s = list(st)
	#a = [int(x) for x in list(s1)]
	#b = [int(x) for x in list(s2)]
	amin = min(a)
	bmin = min(b)
	ans = 0
	for i in range(n):
		ans += max(a[i]-amin,b[i]-bmin)
	print(ans)
		








	




