#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MX = 4e5 + 7;
int n;
long long c[MX][2];
int v[MX];
struct p {
  int x, v;
  inline bool operator<(const p& w) const { return x < w.x; }
} a[MX];
inline int lowbit(int x) { return x & -x; }
void add(int pos, long long k) {
  while (pos <= n) {
    c[pos][0]++;
    c[pos][1] += k;
    pos += lowbit(pos);
  }
}
long long sum(int pos, int k) {
  long long res = 0;
  while (pos) {
    res += c[pos][k];
    pos -= lowbit(pos);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].x);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].v), v[i] = a[i].v;
  sort(a + 1, a + 1 + n);
  sort(v + 1, v + 1 + n);
  int m = unique(v + 1, v + 1 + n) - v - 1;
  long long res = 0;
  for (int i = 1; i <= n; ++i) {
    int pos = lower_bound(v + 1, v + 1 + m, a[i].v) - v;
    res += a[i].x * sum(pos, 0) - sum(pos, 1);
    add(pos, a[i].x);
  }
  printf("%lld\n", res);
  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;
const int INF = 0x3f3f3f3f;
const int MX = 4e5 + 7;
int n;
long long c[MX][2];
int v[MX];
struct p {
  int x, v;
  inline bool operator<(const p& w) const { return x < w.x; }
} a[MX];
inline int lowbit(int x) { return x & -x; }
void add(int pos, long long k) {
  while (pos <= n) {
    c[pos][0]++;
    c[pos][1] += k;
    pos += lowbit(pos);
  }
}
long long sum(int pos, int k) {
  long long res = 0;
  while (pos) {
    res += c[pos][k];
    pos -= lowbit(pos);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].x);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].v), v[i] = a[i].v;
  sort(a + 1, a + 1 + n);
  sort(v + 1, v + 1 + n);
  int m = unique(v + 1, v + 1 + n) - v - 1;
  long long res = 0;
  for (int i = 1; i <= n; ++i) {
    int pos = lower_bound(v + 1, v + 1 + m, a[i].v) - v;
    res += a[i].x * sum(pos, 0) - sum(pos, 1);
    add(pos, a[i].x);
  }
  printf("%lld\n", res);
  return 0;
}

```