#include <bits/stdc++.h>
using namespace std;
int n, T;
pair<int, int> a[200010];
long long getval(vector<long long> &f, int pos) {
  pos++;
  long long ans = 0;
  for (int i = pos; i > 0; i -= (i & -i)) ans += f[i];
  return ans;
}
void upd(vector<long long> &f, int pos, int val) {
  pos++;
  for (int i = pos; i < int(f.size()); i += (i & -i)) f[i] += val;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i].first;
  for (int i = 0; i < n; i++) cin >> a[i].second;
  sort(a, a + n);
  vector<int> compress_v;
  for (int i = 0; i < n; i++) compress_v.push_back(a[i].second);
  sort(compress_v.begin(), compress_v.end());
  compress_v.resize(unique(compress_v.begin(), compress_v.end()) -
                    compress_v.begin());
  vector<long long> cnt(compress_v.size() + 1), sum(compress_v.size() + 1);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int pos = lower_bound(compress_v.begin(), compress_v.end(), a[i].second) -
              compress_v.begin();
    ans += getval(cnt, pos) * 1ll * a[i].first - getval(sum, pos);
    upd(cnt, pos, 1);
    upd(sum, pos, a[i].first);
  }
  cout << ans;
}

### Prompt

Please formulate a CPP solution 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;
int n, T;
pair<int, int> a[200010];
long long getval(vector<long long> &f, int pos) {
  pos++;
  long long ans = 0;
  for (int i = pos; i > 0; i -= (i & -i)) ans += f[i];
  return ans;
}
void upd(vector<long long> &f, int pos, int val) {
  pos++;
  for (int i = pos; i < int(f.size()); i += (i & -i)) f[i] += val;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i].first;
  for (int i = 0; i < n; i++) cin >> a[i].second;
  sort(a, a + n);
  vector<int> compress_v;
  for (int i = 0; i < n; i++) compress_v.push_back(a[i].second);
  sort(compress_v.begin(), compress_v.end());
  compress_v.resize(unique(compress_v.begin(), compress_v.end()) -
                    compress_v.begin());
  vector<long long> cnt(compress_v.size() + 1), sum(compress_v.size() + 1);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int pos = lower_bound(compress_v.begin(), compress_v.end(), a[i].second) -
              compress_v.begin();
    ans += getval(cnt, pos) * 1ll * a[i].first - getval(sum, pos);
    upd(cnt, pos, 1);
    upd(sum, pos, a[i].first);
  }
  cout << ans;
}

```