#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 7;
const int N = 3e5 + 10;
void init(vector<pair<int, int>>& vp) {
  vector<int> v;
  for (auto& p : vp) v.push_back(p.second);
  sort(v.begin(), v.end());
  for (auto& p : vp) {
    p.second = lower_bound(v.begin(), v.end(), p.second) - v.begin() + 1;
  }
}
class BIT {
 public:
  long long a[N];
  void add(int x, long long val) {
    while (x < N) {
      a[x] += val;
      x += x & -x;
    }
  }
  long long sum(int x) {
    long long ret = 0;
    while (x) {
      ret += a[x];
      x -= x & -x;
    }
    return ret;
  }
} b1, b2;
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, int>> vp(n);
  for (auto& p : vp) scanf("%d", &p.first);
  for (auto& p : vp) scanf("%d", &p.second);
  init(vp);
  long long ans = 0;
  sort(vp.begin(), vp.end());
  for (auto& p : vp) {
    b1.add(p.second, 1);
    b2.add(p.second, p.first);
    ans += b1.sum(p.second) * p.first - b2.sum(p.second);
  }
  printf("%lld\n", ans);
  return 0;
}

### 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;
const long long INF = 1e9 + 7;
const int N = 3e5 + 10;
void init(vector<pair<int, int>>& vp) {
  vector<int> v;
  for (auto& p : vp) v.push_back(p.second);
  sort(v.begin(), v.end());
  for (auto& p : vp) {
    p.second = lower_bound(v.begin(), v.end(), p.second) - v.begin() + 1;
  }
}
class BIT {
 public:
  long long a[N];
  void add(int x, long long val) {
    while (x < N) {
      a[x] += val;
      x += x & -x;
    }
  }
  long long sum(int x) {
    long long ret = 0;
    while (x) {
      ret += a[x];
      x -= x & -x;
    }
    return ret;
  }
} b1, b2;
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, int>> vp(n);
  for (auto& p : vp) scanf("%d", &p.first);
  for (auto& p : vp) scanf("%d", &p.second);
  init(vp);
  long long ans = 0;
  sort(vp.begin(), vp.end());
  for (auto& p : vp) {
    b1.add(p.second, 1);
    b2.add(p.second, p.first);
    ans += b1.sum(p.second) * p.first - b2.sum(p.second);
  }
  printf("%lld\n", ans);
  return 0;
}

```