#include <bits/stdc++.h>
using namespace std;
const int N = 200000;
long long BIT1[200001];
long long cnt1[200001];
map<long long, long long> r, l;
pair<long long, long long> arr[200001];
int n;
void upd1(long long idx, long long val) {
  for (; idx <= N; idx = (idx | (idx + 1))) {
    BIT1[idx] += val;
    cnt1[idx]++;
  }
}
pair<long long, long long> query1(int idx) {
  if (idx < 0) return {0, 0};
  long long sum = 0;
  int ct = 0;
  for (; idx >= 0; idx = (idx & (idx + 1)) - 1) {
    sum += BIT1[idx];
    ct += cnt1[idx];
  }
  return {sum, ct};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> arr[i].first;
  for (int i = 1; i <= n; i++) cin >> arr[i].second;
  sort(arr + 1, arr + n + 1);
  for (int i = 1; i <= n; i++) r[arr[i].second] = 0;
  long long ind = 1;
  for (auto &it : r) {
    it.second = ind;
    ind++;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long vel = r[arr[i].second];
    pair<long long, long long> p = query1(vel);
    ans += (p.second) * (arr[i].first);
    ans -= p.first;
    upd1(vel, arr[i].first);
  }
  cout << ans;
}

### 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;
const int N = 200000;
long long BIT1[200001];
long long cnt1[200001];
map<long long, long long> r, l;
pair<long long, long long> arr[200001];
int n;
void upd1(long long idx, long long val) {
  for (; idx <= N; idx = (idx | (idx + 1))) {
    BIT1[idx] += val;
    cnt1[idx]++;
  }
}
pair<long long, long long> query1(int idx) {
  if (idx < 0) return {0, 0};
  long long sum = 0;
  int ct = 0;
  for (; idx >= 0; idx = (idx & (idx + 1)) - 1) {
    sum += BIT1[idx];
    ct += cnt1[idx];
  }
  return {sum, ct};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> arr[i].first;
  for (int i = 1; i <= n; i++) cin >> arr[i].second;
  sort(arr + 1, arr + n + 1);
  for (int i = 1; i <= n; i++) r[arr[i].second] = 0;
  long long ind = 1;
  for (auto &it : r) {
    it.second = ind;
    ind++;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long vel = r[arr[i].second];
    pair<long long, long long> p = query1(vel);
    ans += (p.second) * (arr[i].first);
    ans -= p.first;
    upd1(vel, arr[i].first);
  }
  cout << ans;
}

```