#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
struct Node {
  int x, v;
  bool operator<(const Node &rhs) const { return x < rhs.x; }
} p[maxn];
int b[maxn], n, to;
long long t[maxn], tt[maxn];
inline int lowbit(int x) { return x & -x; }
void update(long long *t, int x, int v) {
  for (; x <= n; x += lowbit(x)) {
    t[x] += v;
  }
}
long long query(long long *t, int x) {
  long long sum = 0;
  for (; x; x -= lowbit(x)) {
    sum += t[x];
  }
  return sum;
}
int find(int x) { return lower_bound(b + 1, b + to + 1, x) - b; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i].x);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &p[i].v);
    b[i] = p[i].v;
  }
  sort(p + 1, p + n + 1);
  sort(b + 1, b + n + 1);
  to = unique(b + 1, b + n + 1) - b - 1;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    int v = find(p[i].v);
    ans += 1LL * query(tt, v) * p[i].x - query(t, v);
    update(t, v, p[i].x);
    update(tt, v, 1);
  }
  printf("%lld\n", ans);
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 maxn = 2e5 + 10;
struct Node {
  int x, v;
  bool operator<(const Node &rhs) const { return x < rhs.x; }
} p[maxn];
int b[maxn], n, to;
long long t[maxn], tt[maxn];
inline int lowbit(int x) { return x & -x; }
void update(long long *t, int x, int v) {
  for (; x <= n; x += lowbit(x)) {
    t[x] += v;
  }
}
long long query(long long *t, int x) {
  long long sum = 0;
  for (; x; x -= lowbit(x)) {
    sum += t[x];
  }
  return sum;
}
int find(int x) { return lower_bound(b + 1, b + to + 1, x) - b; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i].x);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &p[i].v);
    b[i] = p[i].v;
  }
  sort(p + 1, p + n + 1);
  sort(b + 1, b + n + 1);
  to = unique(b + 1, b + n + 1) - b - 1;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    int v = find(p[i].v);
    ans += 1LL * query(tt, v) * p[i].x - query(t, v);
    update(t, v, p[i].x);
    update(tt, v, 1);
  }
  printf("%lld\n", ans);
}

```