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;
bool primecheck(long long int a) {
  bool check = true;
  if (a == 1) return (false);
  for (long long int i = 2; i <= sqrt(a); i++) {
    if (a % i == 0) {
      check = false;
      break;
    }
  }
  return (check);
}
vector<long long int> primeFactors(long long int n) {
  vector<long long int> sol;
  bool once = false;
  while (n % 2 == 0) {
    if (!once) {
      sol.push_back(2);
      once = true;
    }
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    bool once = false;
    while (n % i == 0) {
      if (!once) {
        sol.push_back(i);
        once = true;
      }
      n = n / i;
    }
  }
  if (n > 2) {
    sol.push_back(n);
  }
  return (sol);
}
vector<long long int> printDivisors(long long int n) {
  vector<long long int> sol;
  vector<long long int> another;
  for (long long int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        sol.push_back(i);
      } else {
        sol.push_back(i);
        another.push_back(n / i);
      }
    }
  }
  for (long long int i = another.size() - 1; i >= 0; i--) {
    sol.push_back(another[i]);
  }
  return (sol);
}
long long int digits(long long int a) { return (floor(log10(a)) + 1); }
long long int bits(long long int n) { return __builtin_popcount(n); }
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long int modInverse(long long int a, long long int m) {
  long long int m0 = m;
  long long int y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long int q = a / m;
    long long int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
int gcd(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == b) return a;
  if (a > b) return gcd(a - b, b);
  return gcd(a, b - a);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
vector<long long int> path;
void dfs(long long int u, vector<long long int> adj[], vector<bool> &visited,
         long long int parent = -1) {
  visited[u] = true;
  path[u] = parent;
  for (long long int i = 0; i < adj[u].size(); i++) {
    if (visited[adj[u][i]] == false) {
      dfs(adj[u][i], adj, visited, u);
    }
  }
}
struct Comp {
  bool operator()(const std::pair<int, int> &a, const std::pair<int, int> &b) {
    if (a.first != b.first) {
      return a.first < b.first;
    }
    return a.second > b.second;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int a, b, c, i, j, k, l, m, n, t, p;
  cin >> t;
  while (t--) {
    cin >> a;
    long long int arr[a];
    long long int small1, small2;
    for (i = 0; i < a; i++) {
      cin >> arr[i];
      if (i == 0) {
        small1 = arr[i];
      } else {
        small1 = min(small1, arr[i]);
      }
    }
    long long int second[a];
    for (i = 0; i < a; i++) {
      cin >> second[i];
      if (i == 0) {
        small2 = second[i];
      } else {
        small2 = min(small2, second[i]);
      }
    }
    long long int total = 0;
    for (i = 0; i < a; i++) {
      long long int gap1 = arr[i] - small1;
      long long int gap2 = second[i] - small2;
      long long int small = min(gap1, gap2);
      total += small;
      total += max(gap1, gap2) - small;
    }
    cout << total << "\n";
  }
}
