#include <bits/stdc++.h>
using namespace std;
long long get(vector<long long> &f, int pos) {
  long long res = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) res += f[pos];
  return res;
}
void upd(vector<long long> &f, int pos, int val) {
  for (; pos < int(f.size()); pos |= pos + 1) {
    f[pos] += val;
  }
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int>> p(n);
  for (auto &pnt : p) cin >> pnt.first;
  for (auto &pnt : p) cin >> pnt.second;
  sort(p.begin(), p.end());
  vector<int> vs;
  for (auto &pnt : p) vs.push_back(pnt.second);
  sort(vs.begin(), vs.end());
  vs.resize(unique(vs.begin(), vs.end()) - vs.begin());
  long long ans = 0;
  vector<long long> cnt(vs.size()), xs(vs.size());
  for (auto &pnt : p) {
    int pos = lower_bound(vs.begin(), vs.end(), pnt.second) - vs.begin();
    ans += get(cnt, pos) * 1ll * pnt.first - get(xs, pos);
    upd(cnt, pos, 1);
    upd(xs, pos, pnt.first);
  }
  cout << ans << endl;
  return 0;
}

### 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;
long long get(vector<long long> &f, int pos) {
  long long res = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) res += f[pos];
  return res;
}
void upd(vector<long long> &f, int pos, int val) {
  for (; pos < int(f.size()); pos |= pos + 1) {
    f[pos] += val;
  }
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int>> p(n);
  for (auto &pnt : p) cin >> pnt.first;
  for (auto &pnt : p) cin >> pnt.second;
  sort(p.begin(), p.end());
  vector<int> vs;
  for (auto &pnt : p) vs.push_back(pnt.second);
  sort(vs.begin(), vs.end());
  vs.resize(unique(vs.begin(), vs.end()) - vs.begin());
  long long ans = 0;
  vector<long long> cnt(vs.size()), xs(vs.size());
  for (auto &pnt : p) {
    int pos = lower_bound(vs.begin(), vs.end(), pnt.second) - vs.begin();
    ans += get(cnt, pos) * 1ll * pnt.first - get(xs, pos);
    upd(cnt, pos, 1);
    upd(xs, pos, pnt.first);
  }
  cout << ans << endl;
  return 0;
}

```