#include <bits/stdc++.h>
using namespace std;
vector<long> inverse(vector<long>& perm) {
  long n = perm.size();
  vector<long> ret(n);
  for (long i = 0; i < n; ++i) {
    ret[perm[i]] = i;
  }
  return ret;
}
vector<long> assignPermutation(const vector<long>& uniqs) {
  long n = uniqs.size();
  vector<pair<long, long>> withIndices(n);
  for (long i = 0; i < n; ++i) {
    withIndices[i] = {uniqs[i], i};
  }
  sort(withIndices.begin(), withIndices.end());
  vector<long> temp(n);
  for (long i = 0; i < n; ++i) {
    temp[i] = withIndices[i].second;
  }
  return inverse(temp);
}
int main() {
  long n;
  cin >> n;
  cin.ignore();
  vector<long> x(n);
  vector<long> v(n);
  vector<pair<long, long>> data(n);
  for (long i = 0; i < n; ++i) {
    cin >> x[i];
  }
  cin.ignore();
  for (long i = 0; i < n; ++i) {
    cin >> v[i];
    data[i] = {v[i], x[i]};
  }
  cin.ignore();
  sort(data.begin(), data.end());
  vector<long> xtemp(n);
  for (long i = 0; i < n; ++i) xtemp[i] = data[i].second;
  auto xranks = assignPermutation(xtemp);
  sort(x.begin(), x.end());
  long long totsum = 0;
  for (long i = 0; i < n; ++i) {
    totsum += (long long)(i)*x[i];
    totsum -= (long long)(n - 1 - i) * x[i];
  }
  for (long i = 0; i < n; ++i) {
    long netleftjumps = i - xranks[i];
    totsum += (long long)(netleftjumps)*xtemp[i];
  }
  cout << totsum << endl;
}

### Prompt

Your task is to create a cpp solution to the following problem:

There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.

Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.

Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).

Input

The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.

The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.

The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.

Output

Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).

Examples

Input


3
1 3 2
-100 2 3


Output


3


Input


5
2 1 4 3 5
2 2 2 3 4


Output


19


Input


2
2 1
-3 0


Output


0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<long> inverse(vector<long>& perm) {
  long n = perm.size();
  vector<long> ret(n);
  for (long i = 0; i < n; ++i) {
    ret[perm[i]] = i;
  }
  return ret;
}
vector<long> assignPermutation(const vector<long>& uniqs) {
  long n = uniqs.size();
  vector<pair<long, long>> withIndices(n);
  for (long i = 0; i < n; ++i) {
    withIndices[i] = {uniqs[i], i};
  }
  sort(withIndices.begin(), withIndices.end());
  vector<long> temp(n);
  for (long i = 0; i < n; ++i) {
    temp[i] = withIndices[i].second;
  }
  return inverse(temp);
}
int main() {
  long n;
  cin >> n;
  cin.ignore();
  vector<long> x(n);
  vector<long> v(n);
  vector<pair<long, long>> data(n);
  for (long i = 0; i < n; ++i) {
    cin >> x[i];
  }
  cin.ignore();
  for (long i = 0; i < n; ++i) {
    cin >> v[i];
    data[i] = {v[i], x[i]};
  }
  cin.ignore();
  sort(data.begin(), data.end());
  vector<long> xtemp(n);
  for (long i = 0; i < n; ++i) xtemp[i] = data[i].second;
  auto xranks = assignPermutation(xtemp);
  sort(x.begin(), x.end());
  long long totsum = 0;
  for (long i = 0; i < n; ++i) {
    totsum += (long long)(i)*x[i];
    totsum -= (long long)(n - 1 - i) * x[i];
  }
  for (long i = 0; i < n; ++i) {
    long netleftjumps = i - xranks[i];
    totsum += (long long)(netleftjumps)*xtemp[i];
  }
  cout << totsum << endl;
}

```