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]. 
import java.util.*;
import java.io.* ;
/*
NOTES:
-more than 10 digits (10^10), use long
-prefix sum of 1's and -1's to control flow of highlighting a region
-log(a)/log(b) = log base b of a
-create a position array if trying to see if elements show up next to each other sequentially
 i.e. elements 1,2,3 are within 3 spots of each other in an array
      arr[scannedInt - 1] = i; (position of i at p[i] in original)
- use a freq array when counting instances of an element occurring
- increment the counter first before simulating anything turn-based (two player game)

 */

public class Main {
    static long mod = (long)(Math.pow(10, 9) + 7);
    public static String alpha = "zabcdefghijklmnopqrstuvwxy";
    public static void main(String[] args) throws IOException{
        FastReader in = new FastReader();
        long t= in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            int[] candies = new int[n];
            int[] oranges = new int[n];
            String[] c = in.nextLine().split(" ");
            String[] o = in.nextLine().split(" ");
            for(int i = 0; i<n; i++){
                candies[i] = Integer.parseInt(c[i]);
                oranges[i] = Integer.parseInt(o[i]);
            }
            int minC = Integer.MAX_VALUE;
            int minO = Integer.MAX_VALUE;
            for(int i = 0; i<n; i++){
                minC = Math.min(minC, candies[i]);
                minO = Math.min(minO, oranges[i]);
            }
            long ans = 0;
            for(int i = 0; i<n; i++){
                if(candies[i] == minC && oranges[i] == minO) continue;
                if(candies[i] == minC) ans += oranges[i]-minO;
                else if(oranges[i] == minO) ans += candies[i]-minC;
                else{
                    if(oranges[i]-minO> candies[i]-minC){
                        ans += candies[i]-minC;
                        ans += oranges[i] - (candies[i]-minC) - minO;
                    }
                    else{
                        ans += oranges[i]-minO;
                        ans += candies[i] - (oranges[i]-minO) - minC;
                    }
                }
            }
            System.out.println(ans);
        }

    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e) { e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static ArrayList sieve(){
        ArrayList<Integer> primes = new ArrayList<>();
        int size = 1000001;
        int[] arr = new int[size+1];
        for(int i = 2; i*i<=size; i++){
            if(arr[i] == 0 && (long) (i*i)<=size)
                for(int pp = i*i; pp<=size; pp+=i) arr[pp]++;
        }
        for(int i = 2; i<=size; i++){
            if(arr[i]==0) primes.add(i);
        }
        return primes;
    }


    static boolean isPalindrome(String s){
        for(int i = 0; i<s.length()/2; i++){
            if(s.charAt(i) != s.charAt(s.length()-i-1)) return false;
        }
        return true;
    }

    static ArrayList factors(long n ){
        ArrayList<Long> facts = new ArrayList<>();
        for(int i = 1; i<=(int)Math.sqrt(n); i++){
            if(n%i == 0){
                if( i == Math.sqrt(n)) facts.add((long)i);
                else{
                    facts.add((long)i);
                    facts.add(n/i);
                }
            }
        }
        Collections.sort(facts);
        return facts;
    }

    static boolean isPrime (int n){
        if(factors(n).size() ==2) return true;
        else return false;
    }





    static long GCD(long a, long b) {
        if (b==0) return a;
        return GCD(b,a%b);
    }
    static long LCM(long a, long b){
        return (a*b)/(GCD(a,b));
    }
    static String rev(String s){
        char[] arr = s.toCharArray();
        for(int i = 0; i<(s.length()+1)/2; i++){
            char temp = arr[i];
            arr[i] = arr[s.length()-i-1];
            arr[s.length()-i-1] = temp;
        }
        String fin = new String(arr);
        return fin;
    }
    static long pow(long a, long N) {
        if (N == 0) return 1;
        else if (N == 1) return a;
        else {
            long R = pow(a,N/2);
            if (N % 2 == 0) {
                return R*R;
            }
            else {
                return R*R*a;
            }
        }
    }

    static long powMod(long a, long N) {
        if (N == 0) return 1;
        else if (N == 1) return a % mod;
        else {
            long R = powMod(a,N/2) % mod;
            R *= R;
            R %= mod;
            if (N % 2 == 1) {
                R *= a;
                R %= mod;
            }
            return R % mod;
        }
    }

}

