#include <bits/stdc++.h>
using namespace std;
pair<int, long long> a[200005];
int t[200005];
map<int, int> M;
pair<int, long long> f[200005];
int n;
void update(int gt) {
  int x = M[gt];
  while (x <= n) {
    f[x] = {f[x].first + 1, f[x].second + gt};
    x += (x & -x);
  }
}
pair<int, long long> get(int x) {
  int cnt = 0;
  long long sum = 0;
  while (x >= 1) {
    cnt += f[x].first;
    sum += f[x].second;
    x -= (x & -x);
  }
  return make_pair(cnt, sum);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].second;
    t[i] = a[i].second;
  }
  for (int i = 1; i <= n; i++) cin >> a[i].first;
  sort(a + 1, a + n + 1);
  sort(t + 1, t + n + 1);
  for (int i = 1; i <= n; i++) M[t[i]] = i;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    pair<int, long long> res;
    res = get(M[a[i].second] - 1);
    ans += (1ll * a[i].second * res.first - res.second);
    update(a[i].second);
  }
  cout << ans;
}

### Prompt

Your task is to create 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;
pair<int, long long> a[200005];
int t[200005];
map<int, int> M;
pair<int, long long> f[200005];
int n;
void update(int gt) {
  int x = M[gt];
  while (x <= n) {
    f[x] = {f[x].first + 1, f[x].second + gt};
    x += (x & -x);
  }
}
pair<int, long long> get(int x) {
  int cnt = 0;
  long long sum = 0;
  while (x >= 1) {
    cnt += f[x].first;
    sum += f[x].second;
    x -= (x & -x);
  }
  return make_pair(cnt, sum);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].second;
    t[i] = a[i].second;
  }
  for (int i = 1; i <= n; i++) cin >> a[i].first;
  sort(a + 1, a + n + 1);
  sort(t + 1, t + n + 1);
  for (int i = 1; i <= n; i++) M[t[i]] = i;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    pair<int, long long> res;
    res = get(M[a[i].second] - 1);
    ans += (1ll * a[i].second * res.first - res.second);
    update(a[i].second);
  }
  cout << ans;
}

```