#include <bits/stdc++.h>
using namespace std;
const int mxn = 2e5 + 100;
pair<int, long long> a[mxn];
long long r[mxn], b1[mxn], b2[mxn], ans;
void add(long long *b, int idx, int val, int n) {
  for (; idx <= n; idx += idx & -idx) b[idx] += val;
}
long long ask(long long *b, int idx) {
  long long ret = 0;
  for (; idx > 0; idx -= idx & -idx) ret += b[idx];
  return ret;
}
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].first);
  for (int i = 1; i <= n; i++)
    scanf("%lld", &a[i].second), r[i - 1] = a[i].second;
  sort(r, r + n), m = unique(r, r + n) - r;
  for (int i = 1; i <= n; i++)
    a[i].second = lower_bound(r, r + m, a[i].second) - r + 1;
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    long long x = ask(b2, a[i].second), y = ask(b1, a[i].second);
    ans += 1LL * y * a[i].first - x;
    add(b2, a[i].second, a[i].first, n);
    add(b1, a[i].second, 1, n);
  }
  printf("%lld\n", ans);
}

### Prompt

Please formulate a CPP solution to 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 mxn = 2e5 + 100;
pair<int, long long> a[mxn];
long long r[mxn], b1[mxn], b2[mxn], ans;
void add(long long *b, int idx, int val, int n) {
  for (; idx <= n; idx += idx & -idx) b[idx] += val;
}
long long ask(long long *b, int idx) {
  long long ret = 0;
  for (; idx > 0; idx -= idx & -idx) ret += b[idx];
  return ret;
}
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].first);
  for (int i = 1; i <= n; i++)
    scanf("%lld", &a[i].second), r[i - 1] = a[i].second;
  sort(r, r + n), m = unique(r, r + n) - r;
  for (int i = 1; i <= n; i++)
    a[i].second = lower_bound(r, r + m, a[i].second) - r + 1;
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    long long x = ask(b2, a[i].second), y = ask(b1, a[i].second);
    ans += 1LL * y * a[i].first - x;
    add(b2, a[i].second, a[i].first, n);
    add(b1, a[i].second, 1, n);
  }
  printf("%lld\n", ans);
}

```