#include <bits/stdc++.h>
using namespace std;
long long query(int a, vector<long long>& ft) {
  long long res = 0;
  for (; ~a; a = (a & (a + 1)) - 1) {
    res += ft[a];
  }
  return res;
}
void update(int a, int val, vector<long long>& ft) {
  for (; a < ft.size(); a = a | (a + 1)) {
    ft[a] += val;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> pt(n);
  map<int, int> idx;
  for (auto& i : pt) {
    cin >> i.first;
  }
  for (auto& i : pt) {
    cin >> i.second;
    idx[i.second] = 0;
  }
  sort(pt.begin(), pt.end());
  int sz = 0;
  for (auto& i : idx) {
    i.second = sz++;
  }
  long long ans = 0;
  vector<long long> ct(sz), coord(sz);
  for (auto i : pt) {
    int tmp = idx[i.second];
    ans += i.first * query(tmp, ct) - query(tmp, coord);
    update(tmp, 1, ct);
    update(tmp, i.first, coord);
  }
  cout << ans << '\n';
}

### Prompt

Generate 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;
long long query(int a, vector<long long>& ft) {
  long long res = 0;
  for (; ~a; a = (a & (a + 1)) - 1) {
    res += ft[a];
  }
  return res;
}
void update(int a, int val, vector<long long>& ft) {
  for (; a < ft.size(); a = a | (a + 1)) {
    ft[a] += val;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> pt(n);
  map<int, int> idx;
  for (auto& i : pt) {
    cin >> i.first;
  }
  for (auto& i : pt) {
    cin >> i.second;
    idx[i.second] = 0;
  }
  sort(pt.begin(), pt.end());
  int sz = 0;
  for (auto& i : idx) {
    i.second = sz++;
  }
  long long ans = 0;
  vector<long long> ct(sz), coord(sz);
  for (auto i : pt) {
    int tmp = idx[i.second];
    ans += i.first * query(tmp, ct) - query(tmp, coord);
    update(tmp, 1, ct);
    update(tmp, i.first, coord);
  }
  cout << ans << '\n';
}

```