#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int arr[n];
  int v[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < n; i++) cin >> v[i];
  map<int, int> mp;
  vector<pair<int, int> > vi;
  vector<pair<int, int> > vf;
  for (int i = 0; i < n; i++) {
    vi.push_back({arr[i], v[i]});
    vf.push_back({v[i], arr[i]});
  }
  sort(vi.begin(), vi.end());
  sort(vf.begin(), vf.end());
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int val = vi[i].first;
    mp[val] = i;
    ans += (2 * i - n + 1) * ((long long)val);
  }
  for (int i = 0; i < n; i++) {
    int val = vf[i].second;
    if (mp[val] == i) continue;
    ans += (i - mp[val]) * (long long)val;
  }
  cout << ans << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
int main() {
  int n;
  cin >> n;
  int arr[n];
  int v[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < n; i++) cin >> v[i];
  map<int, int> mp;
  vector<pair<int, int> > vi;
  vector<pair<int, int> > vf;
  for (int i = 0; i < n; i++) {
    vi.push_back({arr[i], v[i]});
    vf.push_back({v[i], arr[i]});
  }
  sort(vi.begin(), vi.end());
  sort(vf.begin(), vf.end());
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int val = vi[i].first;
    mp[val] = i;
    ans += (2 * i - n + 1) * ((long long)val);
  }
  for (int i = 0; i < n; i++) {
    int val = vf[i].second;
    if (mp[val] == i) continue;
    ans += (i - mp[val]) * (long long)val;
  }
  cout << ans << endl;
}

```