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]. 
#include <bits/stdc++.h>
using namespace std;
vector<int> digits(long long num) {
  vector<int> a;
  while (num != 0) {
    a.push_back(num % 10);
    num = num / 10;
  }
  return a;
}
unsigned long long binPower(unsigned long long b, unsigned long long e,
                            unsigned long long m) {
  unsigned long long res = 1;
  b %= m;
  while (e) {
    if (e & 1) res = (unsigned long long)res * b % m;
    b = (unsigned long long)b * b % m;
    e >>= 1;
  }
  return res;
}
bool checkComposite(unsigned long long n, unsigned long long a,
                    unsigned long long d, int s) {
  unsigned long long x = binPower(a, d, n);
  if (x == 1 || x == n - 1) return false;
  for (int r = 1; r < s; r++) {
    x = (unsigned long long)x * x % n;
    if (x == n - 1) return false;
  }
  return true;
}
bool isPrime(unsigned long long n) {
  if (n < 2) return false;
  int r = 0;
  unsigned long long d = n - 1;
  while ((d & 1) == 0) {
    d >>= 1;
    r++;
  }
  int a[12] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
  for (int i = 0; i < 12; ++i) {
    if (n == a[i]) return true;
    if (checkComposite(n, a[i], d, r)) return false;
  }
  return true;
}
void rsolve() {
  unsigned long long n, t, s = 0, al, bl;
  cin >> n;
  vector<long long> a, b;
  for (long long i = 0; i < n; ++i) {
    cin >> t;
    if (i == 0) {
      al = t;
    } else {
      if (al > t) al = t;
    }
    a.push_back(t);
  }
  for (long long i = 0; i < n; ++i) {
    cin >> t;
    if (i == 0) {
      bl = t;
    } else {
      if (bl > t) bl = t;
    }
    b.push_back(t);
  }
  for (long long i = 0; i < n; ++i) {
    if (a[i] > al || b[i] > bl) {
      t = a[i] - al;
      unsigned long long z = b[i] - bl;
      if (t > 0 && z > 0) {
        if (z > t) {
          s += (z - t);
          s += t;
        } else {
          s += z;
          s += (t - z);
        }
      } else {
        s += t;
        s += z;
      }
    }
  }
  cout << s;
}
void solve() {
  int t;
  cin >> t;
  for (int casenum = 1; casenum <= t; ++casenum) {
    rsolve();
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
