#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> o[(int)2e5 + 10];
long long n, ts[(int)2e5 + 10], tc[(int)2e5 + 10];
int main(void) {
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &o[i].second);
  }
  for (int i = 0; i < n; i++) {
    scanf("%lld", &o[i].first);
  }
  sort(o, o + n);
  for (int i = 0; i < n; i++) {
    o[i] = {o[i].second, i + 1};
  }
  sort(o, o + n);
  long long ans = 0;
  for (int i = n - 1; - 1 < i; i--) {
    long long l = o[i].second - 1, r = n, idx = o[i].first;
    long long s = 0, c = 0;
    while (0 < r) {
      s += ts[r];
      c += tc[r];
      r -= (r & -r);
    }
    while (0 < l) {
      s -= ts[l];
      c -= tc[l];
      l -= (l & -l);
    }
    ans += s - c * idx;
    int v = o[i].second;
    while (v <= n) {
      ts[v] += idx;
      tc[v]++;
      v += (v & -v);
    }
  }
  printf("%lld\n", ans);
}

### 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;
pair<long long, long long> o[(int)2e5 + 10];
long long n, ts[(int)2e5 + 10], tc[(int)2e5 + 10];
int main(void) {
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &o[i].second);
  }
  for (int i = 0; i < n; i++) {
    scanf("%lld", &o[i].first);
  }
  sort(o, o + n);
  for (int i = 0; i < n; i++) {
    o[i] = {o[i].second, i + 1};
  }
  sort(o, o + n);
  long long ans = 0;
  for (int i = n - 1; - 1 < i; i--) {
    long long l = o[i].second - 1, r = n, idx = o[i].first;
    long long s = 0, c = 0;
    while (0 < r) {
      s += ts[r];
      c += tc[r];
      r -= (r & -r);
    }
    while (0 < l) {
      s -= ts[l];
      c -= tc[l];
      l -= (l & -l);
    }
    ans += s - c * idx;
    int v = o[i].second;
    while (v <= n) {
      ts[v] += idx;
      tc[v]++;
      v += (v & -v);
    }
  }
  printf("%lld\n", ans);
}

```