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]. 
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**7)
# sys.stdin = open('inpy.txt', 'r')
# sys.stdout = open('outpy.txt', 'w')
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
i_m=9223372036854775807
def inin():
    return int(input())
def stin():
    return input()
def spin():
    return map(int,stin().split())
def lin():                           #takes array as input
    return list(map(int,stin().split()))
def matrix(n):
    #matrix input
    return [list(map(int,input().split()))for i in range(n)]

################################################
def count2Dmatrix(i,list):
    return sum(c.count(i) for c in list)

def modinv(n,p):
    return pow(n,p-2,p)

def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l
prime=[]
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
q=[]       
def dfs(n,d,v,c):
    global q
    v[n]=1
    x=d[n]
    q.append(n)
    j=c
    for i in x:
        if i not in v:
            f=dfs(i,d,v,c+1)
            j=max(j,f)
            # print(f)
    return j
# d = {}
 
"""*******************************************************"""
for _ in range(inin()):
    n = inin()
    a = lin()
    b = lin()
    mina = min(a)
    minb = min(b)

    da, db, c = 0, 0, 0
    
    for i in range(n):
        da = int(math.fabs(mina - a[i]))
        db = int(math.fabs(minb - b[i]))

        c += max(da, db)

    print(c)
