#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
void update(vector<long long>& v, long long ind, long long val) {
  ind++;
  long long n = v.size();
  while (ind <= n) {
    v[ind] += val;
    ind += ((ind) & (-ind));
  }
}
long long query(vector<long long>& v, long long ind) {
  ind++;
  long long res = 0;
  while (ind > 0) {
    res += v[ind];
    ind -= ((ind) & (-ind));
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long, long long>> points(n);
  for (long long i = 0; i < n; i++) cin >> points[i].first;
  vector<long long> vel;
  for (long long i = 0; i < n; i++) {
    cin >> points[i].second;
    vel.push_back(points[i].second);
  }
  sort((vel).begin(), (vel).end());
  sort((points).begin(), (points).end());
  vel.resize(unique((vel).begin(), (vel).end()) - vel.begin());
  vector<long long> no(vel.size() + 1, 0), csum(vel.size() + 1, 0);
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    long long v_new =
        lower_bound(vel.begin(), vel.end(), points[i].second) - vel.begin();
    ans += -(query(csum, v_new)) + (points[i].first * query(no, v_new));
    update(csum, v_new, points[i].first);
    update(no, v_new, 1);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  solve();
}

### Prompt

Please create a solution in CPP 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 INF = 1e9;
void update(vector<long long>& v, long long ind, long long val) {
  ind++;
  long long n = v.size();
  while (ind <= n) {
    v[ind] += val;
    ind += ((ind) & (-ind));
  }
}
long long query(vector<long long>& v, long long ind) {
  ind++;
  long long res = 0;
  while (ind > 0) {
    res += v[ind];
    ind -= ((ind) & (-ind));
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long, long long>> points(n);
  for (long long i = 0; i < n; i++) cin >> points[i].first;
  vector<long long> vel;
  for (long long i = 0; i < n; i++) {
    cin >> points[i].second;
    vel.push_back(points[i].second);
  }
  sort((vel).begin(), (vel).end());
  sort((points).begin(), (points).end());
  vel.resize(unique((vel).begin(), (vel).end()) - vel.begin());
  vector<long long> no(vel.size() + 1, 0), csum(vel.size() + 1, 0);
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    long long v_new =
        lower_bound(vel.begin(), vel.end(), points[i].second) - vel.begin();
    ans += -(query(csum, v_new)) + (points[i].first * query(no, v_new));
    update(csum, v_new, points[i].first);
    update(no, v_new, 1);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  solve();
}

```