#include <bits/stdc++.h>
using namespace std;
int n, c[200005];
long long ans;
struct t {
  long long x, y;
  int z;
} a[200005];
int lowbit(int x) { return x & (-x); }
void update(int x) {
  for (int i = x; i <= n; i += lowbit(i)) c[i]++;
}
int query(int x) {
  int s = 0;
  for (int i = x; i >= 1; i -= lowbit(i)) s += c[i];
  return s;
}
bool cmp(t x, t y) { return x.y < y.y || x.y == y.y && x.x < y.x; }
bool cmp1(t x, t y) { return x.x < y.x; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i].x);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i].y);
  sort(a + 1, a + 1 + n, cmp1);
  for (int i = 1; i <= n; i++) a[i].z = i;
  sort(a + 1, a + 1 + n, cmp);
  memset(c, 0, sizeof(c));
  ans = 0;
  for (int i = 1; i <= n; i++) {
    long long x;
    x = query(a[i].z - 1);
    ans += a[i].x * x;
    update(a[i].z);
  }
  memset(c, 0, sizeof(c));
  for (int i = n; i >= 1; i--) {
    long long x = query(n) - query(a[i].z);
    ans -= a[i].x * x;
    update(a[i].z);
  }
  printf("%lld\n", ans);
}

### 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;
int n, c[200005];
long long ans;
struct t {
  long long x, y;
  int z;
} a[200005];
int lowbit(int x) { return x & (-x); }
void update(int x) {
  for (int i = x; i <= n; i += lowbit(i)) c[i]++;
}
int query(int x) {
  int s = 0;
  for (int i = x; i >= 1; i -= lowbit(i)) s += c[i];
  return s;
}
bool cmp(t x, t y) { return x.y < y.y || x.y == y.y && x.x < y.x; }
bool cmp1(t x, t y) { return x.x < y.x; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i].x);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i].y);
  sort(a + 1, a + 1 + n, cmp1);
  for (int i = 1; i <= n; i++) a[i].z = i;
  sort(a + 1, a + 1 + n, cmp);
  memset(c, 0, sizeof(c));
  ans = 0;
  for (int i = 1; i <= n; i++) {
    long long x;
    x = query(a[i].z - 1);
    ans += a[i].x * x;
    update(a[i].z);
  }
  memset(c, 0, sizeof(c));
  for (int i = n; i >= 1; i--) {
    long long x = query(n) - query(a[i].z);
    ans -= a[i].x * x;
    update(a[i].z);
  }
  printf("%lld\n", ans);
}

```