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]. 
## necessary imports
import sys
input = sys.stdin.readline
# from math import ceil, floor, factorial;

def ceil(x):
    if x != int(x):
        x = int(x) + 1;
    return x;
    
# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp;

## gcd function
def gcd(a,b):
    if b == 0:
        return a;
    return gcd(b, a % b);

## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
    if(k > n - k): 
        k = n - k; 
    res = 1;
    for i in range(k): 
        res = res * (n - i);
        res = res / (i + 1); 
    return int(res);

## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0, hi = None):
    if hi == None:
        hi = len(a);
    while lo < hi:
        mid = (lo+hi)//2;
        if a[mid] < x:
            lo = mid+1;
        else:
            hi = mid;
    return lo;

## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0 and n > 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0 and n > 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes

## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p): 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 

## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b;

# find function with path compression included (recursive)
# def find(x, link):
#     if link[x] == x:
#         return x
#     link[x] = find(link[x], link);
#     return link[x];

# find function with path compression (ITERATIVE)
def find(x, link):
    p = x;
    while( p != link[p]):
        p = link[p];
    
    while( x != p):
        nex = link[x];
        link[x] = p;
        x = nex;
    return p;


# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x

## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    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
    return prime

#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array

################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret;
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

## taking integer array input
def int_array():
    return list(map(int, input().strip().split()));

def float_array():
    return list(map(float, input().strip().split()));

## taking string array input
def str_array():
    return input().strip().split();

#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');

################### ---------------- TEMPLATE ENDS HERE ---------------- ###################

for _ in range(int(input())):
    n = int(input());
    a = int_array(); b = int_array();
    ma = min(a); mb = min(b); ans = 0;
    for i in range(n):
        x = a[i] - ma;
        y = b[i] - mb;
        ans += min(x, y);
        ans += abs(x - y);
    print(ans);