#include <bits/stdc++.h>
using namespace std;
struct BIT {
  vector<long long int> bit;
  int n;
  BIT(long long int N) {
    n = N;
    bit.assign(n + 1, 0);
  }
  void add(int idx, int delta) {
    while (idx <= n) {
      bit[idx] += delta;
      idx += (idx & -idx);
    }
  }
  long long int sum(int idx) {
    long long int res = 0;
    while (idx > 0) {
      res += bit[idx];
      idx -= (idx & -idx);
    }
    return res;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int> > p(n);
  for (auto &i : p) cin >> i.first;
  for (auto &i : p) cin >> i.second;
  sort(p.begin(), p.end());
  vector<int> v;
  for (auto i : p) v.push_back(i.second);
  sort(v.begin(), v.end());
  map<int, int> mp;
  int cnt = 0;
  for (auto i : v) mp[i] = ++cnt;
  for (auto &i : p) i.second = mp[i.second];
  long long int ans = 0;
  BIT tree1(n), tree2(n);
  for (auto i : p) {
    long long int cnt = tree1.sum(i.second);
    long long int sum_x = tree2.sum(i.second);
    ans += cnt * i.first - sum_x;
    tree1.add(i.second, 1);
    tree2.add(i.second, i.first);
  }
  cout << ans << '\n';
}

### Prompt

Create a solution in cpp for 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;
struct BIT {
  vector<long long int> bit;
  int n;
  BIT(long long int N) {
    n = N;
    bit.assign(n + 1, 0);
  }
  void add(int idx, int delta) {
    while (idx <= n) {
      bit[idx] += delta;
      idx += (idx & -idx);
    }
  }
  long long int sum(int idx) {
    long long int res = 0;
    while (idx > 0) {
      res += bit[idx];
      idx -= (idx & -idx);
    }
    return res;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int> > p(n);
  for (auto &i : p) cin >> i.first;
  for (auto &i : p) cin >> i.second;
  sort(p.begin(), p.end());
  vector<int> v;
  for (auto i : p) v.push_back(i.second);
  sort(v.begin(), v.end());
  map<int, int> mp;
  int cnt = 0;
  for (auto i : v) mp[i] = ++cnt;
  for (auto &i : p) i.second = mp[i.second];
  long long int ans = 0;
  BIT tree1(n), tree2(n);
  for (auto i : p) {
    long long int cnt = tree1.sum(i.second);
    long long int sum_x = tree2.sum(i.second);
    ans += cnt * i.first - sum_x;
    tree1.add(i.second, 1);
    tree2.add(i.second, i.first);
  }
  cout << ans << '\n';
}

```