#include <bits/stdc++.h>
using namespace std;
long long query(vector<long long>& ft, int pos) {
  long long ans = 0;
  while (pos > 0) {
    ans += ft[pos - 1];
    pos -= (pos & (-pos));
  }
  return ans;
}
void update(vector<long long>& ft, int pos, long long val, int& ceil) {
  while (pos <= ceil) {
    ft[pos - 1] += val;
    pos += (pos & (-pos));
  }
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int>> p(n);
  vector<int> v(n);
  for (int i = 0; i < n; i++) cin >> p[i].first;
  for (int i = 0; i < n; i++) {
    cin >> p[i].second;
    v[i] = p[i].second;
  }
  sort(p.begin(), p.end());
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  int ceil = v.size();
  vector<long long> ftx(ceil), ftc(ceil);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int pos = lower_bound(v.begin(), v.end(), p[i].second) - v.begin();
    ans += query(ftc, pos + 1) * (1ll * p[i].first) - query(ftx, pos + 1);
    update(ftc, pos + 1, 1, ceil);
    update(ftx, pos + 1, p[i].first, ceil);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
long long query(vector<long long>& ft, int pos) {
  long long ans = 0;
  while (pos > 0) {
    ans += ft[pos - 1];
    pos -= (pos & (-pos));
  }
  return ans;
}
void update(vector<long long>& ft, int pos, long long val, int& ceil) {
  while (pos <= ceil) {
    ft[pos - 1] += val;
    pos += (pos & (-pos));
  }
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int>> p(n);
  vector<int> v(n);
  for (int i = 0; i < n; i++) cin >> p[i].first;
  for (int i = 0; i < n; i++) {
    cin >> p[i].second;
    v[i] = p[i].second;
  }
  sort(p.begin(), p.end());
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  int ceil = v.size();
  vector<long long> ftx(ceil), ftc(ceil);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int pos = lower_bound(v.begin(), v.end(), p[i].second) - v.begin();
    ans += query(ftc, pos + 1) * (1ll * p[i].first) - query(ftx, pos + 1);
    update(ftc, pos + 1, 1, ceil);
    update(ftx, pos + 1, p[i].first, ceil);
  }
  cout << ans << endl;
  return 0;
}

```