#include <bits/stdc++.h>
using namespace std;
long long bit1[200005], bit2[200005];
inline void upd(long long idx, long long val, long long bit[]) {
  while (idx < 200005) {
    bit[idx] += val;
    idx += (idx & -idx);
  }
}
inline long long que(long long idx, long long bit[]) {
  long long ans = 0;
  while (idx > 0) {
    ans += bit[idx];
    idx -= (idx & -idx);
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> v;
  map<long long, long long> m;
  pair<long long, long long> arr[n];
  for (long long i = 0; i < n; i++) cin >> arr[i].first;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i].second;
    if (m[arr[i].second] == 0) {
      m[arr[i].second]++;
      v.push_back(arr[i].second);
    }
  }
  sort(v.begin(), v.end());
  sort(arr, arr + n);
  for (long long i = 0; i < n; i++) {
    long long x = lower_bound(v.begin(), v.end(), arr[i].second) - v.begin();
    arr[i].second = x + 1;
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    long long x = que(arr[i].second, bit1), y = que(arr[i].second, bit2);
    ans += (x * arr[i].first - y);
    upd(arr[i].second, 1, bit1);
    upd(arr[i].second, arr[i].first, bit2);
  }
  cout << ans << '\n';
}

### 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;
long long bit1[200005], bit2[200005];
inline void upd(long long idx, long long val, long long bit[]) {
  while (idx < 200005) {
    bit[idx] += val;
    idx += (idx & -idx);
  }
}
inline long long que(long long idx, long long bit[]) {
  long long ans = 0;
  while (idx > 0) {
    ans += bit[idx];
    idx -= (idx & -idx);
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> v;
  map<long long, long long> m;
  pair<long long, long long> arr[n];
  for (long long i = 0; i < n; i++) cin >> arr[i].first;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i].second;
    if (m[arr[i].second] == 0) {
      m[arr[i].second]++;
      v.push_back(arr[i].second);
    }
  }
  sort(v.begin(), v.end());
  sort(arr, arr + n);
  for (long long i = 0; i < n; i++) {
    long long x = lower_bound(v.begin(), v.end(), arr[i].second) - v.begin();
    arr[i].second = x + 1;
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    long long x = que(arr[i].second, bit1), y = que(arr[i].second, bit2);
    ans += (x * arr[i].first - y);
    upd(arr[i].second, 1, bit1);
    upd(arr[i].second, arr[i].first, bit2);
  }
  cout << ans << '\n';
}

```