#include <bits/stdc++.h>
using namespace std;
long long int get(vector<long long int> &arr, long long int ind,
                  long long int n) {
  long long int ans = 0;
  for (; ind > 0; ind -= (ind & (-ind))) ans += arr[ind];
  return ans;
}
void upd(vector<long long int> &arr, long long int ind, long long int x,
         long long int n) {
  for (; ind <= n; ind += (ind & (-ind))) arr[ind] += x;
}
int main() {
  long long int n;
  cin >> n;
  pair<long long int, long long int> arr[n];
  for (long long int i = 0; i < n; i++) cin >> arr[i].first;
  for (long long int i = 0; i < n; i++) cin >> arr[i].second;
  vector<int> vec(n);
  for (long long int i = 0; i < n; i++) vec.push_back(arr[i].second);
  sort(vec.begin(), vec.end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  vector<long long int> cnt(vec.size() + 1), cs(vec.size() + 1);
  sort(arr, arr + n);
  long long int ans = 0;
  long long int n1 = vec.size();
  for (long long int i = 0; i < n; i++) {
    long long int pos =
        lower_bound(vec.begin(), vec.end(), arr[i].second) - vec.begin();
    ans += get(cnt, pos + 1, n1) * arr[i].first - get(cs, pos + 1, n1);
    upd(cnt, pos + 1, 1, n1);
    upd(cs, pos + 1, arr[i].first, n1);
  }
  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 int get(vector<long long int> &arr, long long int ind,
                  long long int n) {
  long long int ans = 0;
  for (; ind > 0; ind -= (ind & (-ind))) ans += arr[ind];
  return ans;
}
void upd(vector<long long int> &arr, long long int ind, long long int x,
         long long int n) {
  for (; ind <= n; ind += (ind & (-ind))) arr[ind] += x;
}
int main() {
  long long int n;
  cin >> n;
  pair<long long int, long long int> arr[n];
  for (long long int i = 0; i < n; i++) cin >> arr[i].first;
  for (long long int i = 0; i < n; i++) cin >> arr[i].second;
  vector<int> vec(n);
  for (long long int i = 0; i < n; i++) vec.push_back(arr[i].second);
  sort(vec.begin(), vec.end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  vector<long long int> cnt(vec.size() + 1), cs(vec.size() + 1);
  sort(arr, arr + n);
  long long int ans = 0;
  long long int n1 = vec.size();
  for (long long int i = 0; i < n; i++) {
    long long int pos =
        lower_bound(vec.begin(), vec.end(), arr[i].second) - vec.begin();
    ans += get(cnt, pos + 1, n1) * arr[i].first - get(cs, pos + 1, n1);
    upd(cnt, pos + 1, 1, n1);
    upd(cs, pos + 1, arr[i].first, n1);
  }
  cout << ans << endl;
  return 0;
}

```