#include <bits/stdc++.h>
using namespace std;
struct Node {
  int X, V;
  inline friend bool operator<(Node x, Node y) {
    if (x.V == y.V) {
      return x.X < y.X;
    }
    return x.V < y.V;
  }
} p[200000];
map<int, int> A;
set<int> B;
int ct;
long long c1[200001];
int c2[200001];
inline void Add(int x, int d) {
  for (register int i = x; i <= ct; i += i & -i) {
    c1[i] += d;
    c2[i]++;
  }
}
inline pair<long long, int> Get(int x) {
  int s2 = 0;
  long long s1 = 0;
  for (register int i = x; i != 0; i -= i & -i) {
    s1 += c1[i];
    s2 += c2[i];
  }
  return make_pair(s1, s2);
}
int main() {
  int n, k;
  scanf("%d", &n);
  for (register int i = 0; i != n; i++) {
    scanf("%d", &p[i].X);
    B.insert(p[i].X);
  }
  for (register int i = 0; i != n; i++) {
    scanf("%d", &p[i].V);
  }
  sort(p, p + n);
  for (set<int>::iterator T = B.begin(); T != B.end(); T++) {
    ct++;
    A[*T] = ct;
  }
  long long ans = 0;
  for (register int i = 0; i != n; i++) {
    k = A[p[i].X];
    pair<long long, int> Tem = Get(k);
    ans += (long long)p[i].X * Tem.second - Tem.first;
    Add(k, p[i].X);
  }
  printf("%lld", ans);
  return 0;
}

### 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;
struct Node {
  int X, V;
  inline friend bool operator<(Node x, Node y) {
    if (x.V == y.V) {
      return x.X < y.X;
    }
    return x.V < y.V;
  }
} p[200000];
map<int, int> A;
set<int> B;
int ct;
long long c1[200001];
int c2[200001];
inline void Add(int x, int d) {
  for (register int i = x; i <= ct; i += i & -i) {
    c1[i] += d;
    c2[i]++;
  }
}
inline pair<long long, int> Get(int x) {
  int s2 = 0;
  long long s1 = 0;
  for (register int i = x; i != 0; i -= i & -i) {
    s1 += c1[i];
    s2 += c2[i];
  }
  return make_pair(s1, s2);
}
int main() {
  int n, k;
  scanf("%d", &n);
  for (register int i = 0; i != n; i++) {
    scanf("%d", &p[i].X);
    B.insert(p[i].X);
  }
  for (register int i = 0; i != n; i++) {
    scanf("%d", &p[i].V);
  }
  sort(p, p + n);
  for (set<int>::iterator T = B.begin(); T != B.end(); T++) {
    ct++;
    A[*T] = ct;
  }
  long long ans = 0;
  for (register int i = 0; i != n; i++) {
    k = A[p[i].X];
    pair<long long, int> Tem = Get(k);
    ans += (long long)p[i].X * Tem.second - Tem.first;
    Add(k, p[i].X);
  }
  printf("%lld", ans);
  return 0;
}

```