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;
using namespace std;
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long gcd(long long a, long long b) {
  if (a < b)
    return gcd(b, a);
  else if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long isPrime(long long n) {
  long long p = (long long)sqrt(n);
  for (long long i = 2; 2 < p + 1 ? i < p + 1 : i > p + 1;
       2 < p + 1 ? i += 1 : i -= 1)
    if (n % i == 0) return 0;
  return 1;
}
long long pow(long long b, long long e) {
  if (e == 0)
    return 1;
  else if (e % 2 == 0) {
    long long a = pow(b, e / 2);
    return a * a;
  } else {
    long long a = pow(b, e / 2);
    return b * a * a;
  }
}
long long pow_m(long long x, long long y, long long m = 1000000007) {
  x = x % m;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= m;
    y = y >> 1;
    x = x * x;
    x %= m;
  }
  return res;
}
long long zeros(long long n) {
  long long cnt = 0;
  while (n) {
    n /= 5;
    cnt += n;
  }
  return cnt;
}
long long modInverse(long long a, long long m) { return pow_m(a, m - 2, m); }
long long max(long long a, long long b) {
  if (a >= b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a <= b)
    return a;
  else
    return b;
}
long long sum(vector<long long> v) {
  long long sm = 0;
  for (long long i = 0; 0 < v.size() ? i < v.size() : i > v.size();
       0 < v.size() ? i += 1 : i -= 1)
    sm += v[i];
  return sm;
}
long long bin_coff(long long n, long long k, long long m) {
  vector<vector<long long> > ans(n + 1, vector<long long>(k, 0));
  for (long long i = 0; 0 < n + 1 ? i < n + 1 : i > n + 1;
       0 < n + 1 ? i += 1 : i -= 1) {
    for (long long j = 0;
         0 < min(i, k) + 1 ? j < min(i, k) + 1 : j > min(i, k) + 1;
         0 < min(i, k) + 1 ? j += 1 : j -= 1) {
      if (j == 0 || j == i)
        ans[i][j] = 1;
      else
        ans[i][j] = ans[i - 1][j - 1] % m + ans[i - 1][j] % m;
    }
  }
  return ans[n][k] % m;
}
long long inverse(long long i) {
  if (i == 1) return 1;
  return (1000000007 -
          ((1000000007 / i) * inverse(1000000007 % i)) % 1000000007 +
          1000000007) %
         1000000007;
}
long long maxProduct(vector<long long> arr, long long n, long long k) {
  sort(arr.begin(), arr.end());
  long long prod = 1;
  if (arr[n - 1] == 0 && (k & 1)) return 0;
  if (arr[n - 1] <= 0 && (k & 1)) {
    for (long long i = n - 1; i >= n - k; i--) prod *= arr[i];
    return prod;
  }
  long long i = 0, j = n - 1;
  if (k & 1) {
    prod *= arr[j];
    j--;
    k--;
  }
  k >>= 1;
  for (long long itr = 0; itr < k; itr++) {
    long long left_product = arr[i] * arr[i + 1];
    long long right_product = arr[j] * arr[j - 1];
    if (left_product > right_product) {
      prod *= left_product;
      i += 2;
    } else {
      prod *= right_product;
      j -= 2;
    }
  }
  return prod;
}
void solve() {
  long long n;
  cin >> n;
  long long minn1 = 1e18, minn2 = 1e18;
  vector<long long> a(n), b(n);
  for (long long i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1) {
    cin >> a[i];
    minn1 = min(minn1, a[i]);
  }
  for (long long i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1) {
    cin >> b[i];
    minn2 = min(minn2, b[i]);
  }
  long long cnt = 0;
  for (long long i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1) {
    cnt += max(a[i] - minn1, b[i] - minn2);
  }
  cout << cnt << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long T;
  cin >> T;
  while (T--) solve();
  return 0;
}
