#include <bits/stdc++.h>
using namespace std;
bool test_cases_exist = false;
class FenwickTree {
 public:
  vector<long long int> tree;
  long long int n;
  FenwickTree(long long int n) {
    this->n = n;
    tree.resize(n + 5);
  }
  void update(long long int ind, long long int delta) {
    while (ind <= n) {
      tree[ind] += delta;
      ind += ind & -ind;
    }
  }
  long long int query(long long int ind) {
    long long int res = 0;
    while (ind >= 1) {
      res += tree[ind];
      ind -= ind & -ind;
    }
    return res;
  }
};
void solve() {
  long long int n;
  cin >> n;
  vector<pair<long long int, long long int>> pnt(n);
  for (auto &z : pnt) cin >> z.first;
  for (auto &z : pnt) cin >> z.second;
  vector<long long int> speed;
  for (auto &z : pnt) speed.push_back(z.second);
  sort(pnt.begin(), pnt.end());
  sort(speed.begin(), speed.end());
  long long int ans = 0;
  vector<FenwickTree> ft(2, {n});
  for (auto z : pnt) {
    int pos = lower_bound(speed.begin(), speed.end(), z.second) - speed.begin();
    ans += ft[0].query(pos + 1) * z.first;
    ans -= ft[1].query(pos + 1);
    ft[0].update(pos + 1, 1);
    ft[1].update(pos + 1, z.first);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  if (test_cases_exist) cin >> t;
  while (t--) solve();
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
bool test_cases_exist = false;
class FenwickTree {
 public:
  vector<long long int> tree;
  long long int n;
  FenwickTree(long long int n) {
    this->n = n;
    tree.resize(n + 5);
  }
  void update(long long int ind, long long int delta) {
    while (ind <= n) {
      tree[ind] += delta;
      ind += ind & -ind;
    }
  }
  long long int query(long long int ind) {
    long long int res = 0;
    while (ind >= 1) {
      res += tree[ind];
      ind -= ind & -ind;
    }
    return res;
  }
};
void solve() {
  long long int n;
  cin >> n;
  vector<pair<long long int, long long int>> pnt(n);
  for (auto &z : pnt) cin >> z.first;
  for (auto &z : pnt) cin >> z.second;
  vector<long long int> speed;
  for (auto &z : pnt) speed.push_back(z.second);
  sort(pnt.begin(), pnt.end());
  sort(speed.begin(), speed.end());
  long long int ans = 0;
  vector<FenwickTree> ft(2, {n});
  for (auto z : pnt) {
    int pos = lower_bound(speed.begin(), speed.end(), z.second) - speed.begin();
    ans += ft[0].query(pos + 1) * z.first;
    ans -= ft[1].query(pos + 1);
    ft[0].update(pos + 1, 1);
    ft[1].update(pos + 1, z.first);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  if (test_cases_exist) cin >> t;
  while (t--) solve();
}

```