#include <bits/stdc++.h>
int xx[200000], vv[200000];
int compare_v(const void *a, const void *b) {
  int i = *(int *)a;
  int j = *(int *)b;
  return vv[i] - vv[j];
}
int compare_x(const void *a, const void *b) {
  int i = *(int *)a;
  int j = *(int *)b;
  return xx[i] - xx[j];
}
void sort(int *ii, int n, int (*compare)(const void *, const void *)) {
  qsort(ii, n, sizeof *ii, compare);
}
void update(long long *ft, int i, int n, int x) {
  while (i < n) {
    ft[i] += x;
    i |= i + 1;
  }
}
long long query(long long *ft, int i) {
  long long x = 0;
  while (i >= 0) {
    x += ft[i];
    i &= i + 1, i--;
  }
  return x;
}
int main() {
  static long long ft_cnt[200000], ft_sum[200000];
  static int ii[200000];
  int n, n_, i;
  long long ans;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &xx[i]);
  for (i = 0; i < n; i++) scanf("%d", &vv[i]);
  for (i = 0; i < n; i++) ii[i] = i;
  sort(ii, n, compare_v);
  n_ = 0;
  for (i = 0; i < n; i++)
    vv[ii[i]] = i + 1 == n || vv[ii[i + 1]] != vv[ii[i]] ? n_++ : n_;
  sort(ii, n, compare_x);
  ans = 0;
  for (i = 0; i < n; i++) {
    int i_ = ii[i], x = xx[i_], v = vv[i_];
    ans += query(ft_cnt, v) * x - query(ft_sum, v);
    update(ft_cnt, v, n_, 1);
    update(ft_sum, v, n_, x);
  }
  printf("%lld\n", ans);
  return 0;
}

### 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>
int xx[200000], vv[200000];
int compare_v(const void *a, const void *b) {
  int i = *(int *)a;
  int j = *(int *)b;
  return vv[i] - vv[j];
}
int compare_x(const void *a, const void *b) {
  int i = *(int *)a;
  int j = *(int *)b;
  return xx[i] - xx[j];
}
void sort(int *ii, int n, int (*compare)(const void *, const void *)) {
  qsort(ii, n, sizeof *ii, compare);
}
void update(long long *ft, int i, int n, int x) {
  while (i < n) {
    ft[i] += x;
    i |= i + 1;
  }
}
long long query(long long *ft, int i) {
  long long x = 0;
  while (i >= 0) {
    x += ft[i];
    i &= i + 1, i--;
  }
  return x;
}
int main() {
  static long long ft_cnt[200000], ft_sum[200000];
  static int ii[200000];
  int n, n_, i;
  long long ans;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &xx[i]);
  for (i = 0; i < n; i++) scanf("%d", &vv[i]);
  for (i = 0; i < n; i++) ii[i] = i;
  sort(ii, n, compare_v);
  n_ = 0;
  for (i = 0; i < n; i++)
    vv[ii[i]] = i + 1 == n || vv[ii[i + 1]] != vv[ii[i]] ? n_++ : n_;
  sort(ii, n, compare_x);
  ans = 0;
  for (i = 0; i < n; i++) {
    int i_ = ii[i], x = xx[i_], v = vv[i_];
    ans += query(ft_cnt, v) * x - query(ft_sum, v);
    update(ft_cnt, v, n_, 1);
    update(ft_sum, v, n_, x);
  }
  printf("%lld\n", ans);
  return 0;
}

```