#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
pair<int, int> p[maxn];
int n;
long long d[maxn];
long long e[maxn];
void add(long long *c, int i, int k) {
  while (i <= n) {
    c[i] += k;
    i += i & (-i);
  }
}
long long sum(long long *c, int i) {
  long long res = 0;
  while (i > 0) {
    res += c[i];
    i -= i & (-i);
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 0; i <= n - 1; i++) cin >> p[i].first;
  for (int i = 0; i <= n - 1; i++) cin >> p[i].second;
  sort(p, p + n);
  vector<int> es(n);
  for (int i = 0; i <= n - 1; i++) es[i] = p[i].second;
  sort(es.begin(), es.end());
  long long res = 0;
  for (int i = 0; i <= n - 1; i++) {
    int pp = lower_bound(es.begin(), es.end(), p[i].second) - es.begin() + 1;
    res += sum(d, pp) * p[i].first - sum(e, pp);
    add(d, pp, 1);
    add(e, pp, p[i].first);
  }
  cout << res << "\n";
}

### 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;
const int maxn = 1e6 + 5;
pair<int, int> p[maxn];
int n;
long long d[maxn];
long long e[maxn];
void add(long long *c, int i, int k) {
  while (i <= n) {
    c[i] += k;
    i += i & (-i);
  }
}
long long sum(long long *c, int i) {
  long long res = 0;
  while (i > 0) {
    res += c[i];
    i -= i & (-i);
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 0; i <= n - 1; i++) cin >> p[i].first;
  for (int i = 0; i <= n - 1; i++) cin >> p[i].second;
  sort(p, p + n);
  vector<int> es(n);
  for (int i = 0; i <= n - 1; i++) es[i] = p[i].second;
  sort(es.begin(), es.end());
  long long res = 0;
  for (int i = 0; i <= n - 1; i++) {
    int pp = lower_bound(es.begin(), es.end(), p[i].second) - es.begin() + 1;
    res += sum(d, pp) * p[i].first - sum(e, pp);
    add(d, pp, 1);
    add(e, pp, p[i].first);
  }
  cout << res << "\n";
}

```