#include <bits/stdc++.h>
using namespace std;
int n;
long long BIT[515151];
int BIT2[515151];
map<int, int> hs;
void update(int x, int y) {
  for (int i = x; i <= 5e5; i += (i & (-i))) BIT[i] += y, BIT2[i]++;
}
long long query(int x) {
  long long tmp = 0;
  for (int i = x; i >= 1; i -= (i & (-i))) tmp += BIT[i];
  return tmp;
}
long long query2(int x) {
  int tmp = 0;
  for (int i = x; i >= 1; i -= (i & (-i))) tmp += BIT2[i];
  return tmp;
}
int main() {
  scanf("%d", &n);
  int x[n + 5], v[n + 5];
  for (int i = 0; i < n; i++) {
    scanf("%d", &x[i]);
  }
  vector<int> tmp;
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
    tmp.push_back(v[i]);
  }
  sort(tmp.begin(), tmp.end());
  int idx = 1;
  for (auto i : tmp) {
    hs[i] = idx;
    idx++;
  }
  for (int i = 0; i < n; i++) {
    v[i] = hs[v[i]];
  }
  vector<pair<int, int> > data;
  for (int i = 0; i < n; i++) {
    data.push_back(make_pair(x[i], v[i]));
  }
  sort(data.begin(), data.end());
  long long ans = 0;
  for (auto i : data) {
    long long cnt1 = query(i.second);
    long long cnt2 = query2(i.second);
    ans += (cnt2 * i.first - cnt1);
    update(i.second, i.first);
  }
  printf("%I64d\n", ans);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long BIT[515151];
int BIT2[515151];
map<int, int> hs;
void update(int x, int y) {
  for (int i = x; i <= 5e5; i += (i & (-i))) BIT[i] += y, BIT2[i]++;
}
long long query(int x) {
  long long tmp = 0;
  for (int i = x; i >= 1; i -= (i & (-i))) tmp += BIT[i];
  return tmp;
}
long long query2(int x) {
  int tmp = 0;
  for (int i = x; i >= 1; i -= (i & (-i))) tmp += BIT2[i];
  return tmp;
}
int main() {
  scanf("%d", &n);
  int x[n + 5], v[n + 5];
  for (int i = 0; i < n; i++) {
    scanf("%d", &x[i]);
  }
  vector<int> tmp;
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
    tmp.push_back(v[i]);
  }
  sort(tmp.begin(), tmp.end());
  int idx = 1;
  for (auto i : tmp) {
    hs[i] = idx;
    idx++;
  }
  for (int i = 0; i < n; i++) {
    v[i] = hs[v[i]];
  }
  vector<pair<int, int> > data;
  for (int i = 0; i < n; i++) {
    data.push_back(make_pair(x[i], v[i]));
  }
  sort(data.begin(), data.end());
  long long ans = 0;
  for (auto i : data) {
    long long cnt1 = query(i.second);
    long long cnt2 = query2(i.second);
    ans += (cnt2 * i.first - cnt1);
    update(i.second, i.first);
  }
  printf("%I64d\n", ans);
}

```