#include <bits/stdc++.h>
using namespace std;
const int N = 200000;
int x[N + 5], v[N + 5], _v[N + 5];
pair<int, int> point[N + 5];
template <typename T, int N>
struct Binary_Indexed_Tree {
  int n;
  T sumv[N + 5];
  void init(int _n) {
    n = _n;
    memset(sumv, 0, (n + 4) * sizeof(T));
  }
  void build(T A[], int _n) {
    init(_n);
    for (int i = 1; i <= n; ++i) {
      sumv[i] += A[i];
      int j = i + (i & -i);
      if (j <= n) sumv[j] += sumv[i];
    }
  }
  void add(int p, T k) {
    while (p <= n) {
      sumv[p] += k;
      p += p & -p;
    }
  }
  T sum(int p) {
    T res = 0;
    while (p) {
      res += sumv[p];
      p -= p & -p;
    }
    return res;
  }
};
Binary_Indexed_Tree<long long, N> bit, bitr, cbit, cbitr;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", x + i);
  for (int i = 1; i <= n; ++i) scanf("%d", v + i);
  memcpy(_v, v, sizeof(v));
  sort(_v + 1, _v + n + 1);
  auto ed = unique(_v + 1, _v + n + 1);
  for (int i = 1; i <= n; ++i) {
    v[i] = lower_bound(_v + 1, ed, v[i]) - _v;
    point[i] = {x[i], v[i]};
  }
  sort(point + 1, point + n + 1);
  bit.init(n);
  bitr.init(n);
  cbit.init(n);
  cbitr.init(n);
  for (int i = 1; i <= n; ++i) {
    int xi = point[i].first, vi = point[i].second;
    bitr.add(vi, xi);
    cbitr.add(vi, 1);
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    int xi = point[i].first, vi = point[i].second;
    bitr.add(vi, -xi);
    cbitr.add(vi, -1);
    ans += xi * cbit.sum(vi) - bit.sum(vi) + bitr.sum(n) - bitr.sum(vi - 1) -
           (cbitr.sum(n) - cbitr.sum(vi - 1)) * xi;
    bit.add(vi, xi);
    cbit.add(vi, 1);
  }
  printf("%lld\n", ans / 2);
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int x[N + 5], v[N + 5], _v[N + 5];
pair<int, int> point[N + 5];
template <typename T, int N>
struct Binary_Indexed_Tree {
  int n;
  T sumv[N + 5];
  void init(int _n) {
    n = _n;
    memset(sumv, 0, (n + 4) * sizeof(T));
  }
  void build(T A[], int _n) {
    init(_n);
    for (int i = 1; i <= n; ++i) {
      sumv[i] += A[i];
      int j = i + (i & -i);
      if (j <= n) sumv[j] += sumv[i];
    }
  }
  void add(int p, T k) {
    while (p <= n) {
      sumv[p] += k;
      p += p & -p;
    }
  }
  T sum(int p) {
    T res = 0;
    while (p) {
      res += sumv[p];
      p -= p & -p;
    }
    return res;
  }
};
Binary_Indexed_Tree<long long, N> bit, bitr, cbit, cbitr;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", x + i);
  for (int i = 1; i <= n; ++i) scanf("%d", v + i);
  memcpy(_v, v, sizeof(v));
  sort(_v + 1, _v + n + 1);
  auto ed = unique(_v + 1, _v + n + 1);
  for (int i = 1; i <= n; ++i) {
    v[i] = lower_bound(_v + 1, ed, v[i]) - _v;
    point[i] = {x[i], v[i]};
  }
  sort(point + 1, point + n + 1);
  bit.init(n);
  bitr.init(n);
  cbit.init(n);
  cbitr.init(n);
  for (int i = 1; i <= n; ++i) {
    int xi = point[i].first, vi = point[i].second;
    bitr.add(vi, xi);
    cbitr.add(vi, 1);
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    int xi = point[i].first, vi = point[i].second;
    bitr.add(vi, -xi);
    cbitr.add(vi, -1);
    ans += xi * cbit.sum(vi) - bit.sum(vi) + bitr.sum(n) - bitr.sum(vi - 1) -
           (cbitr.sum(n) - cbitr.sum(vi - 1)) * xi;
    bit.add(vi, xi);
    cbit.add(vi, 1);
  }
  printf("%lld\n", ans / 2);
  return 0;
}

```