#include <bits/stdc++.h>
using rmain = int;
using namespace std;
long long n;
vector<pair<long long, long long>> points;
vector<long long> v, cnt, xs;
long long LSOne(long long k) { return (k & (-k)); }
void update(vector<long long>& f, long long pos, long long val) {
  for (; pos <= n; pos += LSOne(pos)) f[pos] += val;
}
long long rsq(vector<long long>& f, long long pos) {
  long long sum = 0;
  for (; pos; pos -= LSOne(pos)) sum += f[pos];
  return sum;
}
rmain main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  points.resize(n + 1);
  v.resize(n + 1);
  cnt.resize(n + 1);
  xs.resize(n + 1);
  for (long long i = 1; i <= n; ++i) cin >> points[i].first;
  for (long long i = 1; i <= n; ++i) {
    cin >> points[i].second;
    v[i] = points[i].second;
  }
  sort(points.begin() + 1, points.begin() + n + 1);
  sort(v.begin() + 1, v.begin() + n + 1);
  long long ans = 0;
  for (long long i = 1; i <= n; ++i) {
    long long pos =
        lower_bound(v.begin() + 1, v.begin() + n + 1, points[i].second) -
        v.begin();
    ans += rsq(cnt, pos) * points[i].first - rsq(xs, pos);
    update(cnt, pos, 1);
    update(xs, pos, points[i].first);
  }
  cout << ans;
}

### Prompt

Generate 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 rmain = int;
using namespace std;
long long n;
vector<pair<long long, long long>> points;
vector<long long> v, cnt, xs;
long long LSOne(long long k) { return (k & (-k)); }
void update(vector<long long>& f, long long pos, long long val) {
  for (; pos <= n; pos += LSOne(pos)) f[pos] += val;
}
long long rsq(vector<long long>& f, long long pos) {
  long long sum = 0;
  for (; pos; pos -= LSOne(pos)) sum += f[pos];
  return sum;
}
rmain main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  points.resize(n + 1);
  v.resize(n + 1);
  cnt.resize(n + 1);
  xs.resize(n + 1);
  for (long long i = 1; i <= n; ++i) cin >> points[i].first;
  for (long long i = 1; i <= n; ++i) {
    cin >> points[i].second;
    v[i] = points[i].second;
  }
  sort(points.begin() + 1, points.begin() + n + 1);
  sort(v.begin() + 1, v.begin() + n + 1);
  long long ans = 0;
  for (long long i = 1; i <= n; ++i) {
    long long pos =
        lower_bound(v.begin() + 1, v.begin() + n + 1, points[i].second) -
        v.begin();
    ans += rsq(cnt, pos) * points[i].first - rsq(xs, pos);
    update(cnt, pos, 1);
    update(xs, pos, points[i].first);
  }
  cout << ans;
}

```