#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long a = 0, b = getchar(), c = 1;
  while (!isdigit(b)) c = b == '-' ? -1 : 1, b = getchar();
  while (isdigit(b)) a = a * 10 + b - '0', b = getchar();
  return a * c;
}
long long n, ans, p[200005];
pair<long long, long long> a[200005];
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i].second = read();
    p[i] = a[i].second;
  }
  for (int i = 1; i <= n; i++) a[i].first = read();
  sort(a + 1, a + n + 1);
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; i++)
    ans += (lower_bound(p + 1, p + n + 1, a[i].second) - p - n + i - 1) *
           a[i].second;
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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 read() {
  long long a = 0, b = getchar(), c = 1;
  while (!isdigit(b)) c = b == '-' ? -1 : 1, b = getchar();
  while (isdigit(b)) a = a * 10 + b - '0', b = getchar();
  return a * c;
}
long long n, ans, p[200005];
pair<long long, long long> a[200005];
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i].second = read();
    p[i] = a[i].second;
  }
  for (int i = 1; i <= n; i++) a[i].first = read();
  sort(a + 1, a + n + 1);
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; i++)
    ans += (lower_bound(p + 1, p + n + 1, a[i].second) - p - n + i - 1) *
           a[i].second;
  cout << ans;
  return 0;
}

```