#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5;
int N;
pair<long long, long long> A[MAXN + 10];
long long ans;
vector<long long> comp;
long long tree1[MAXN + 10];
void update1(int i, long long v) {
  for (; i <= N; i += (i & -i)) tree1[i] += v;
}
long long query1(int i) {
  long long ret = 0;
  for (; i > 0; i -= (i & -i)) ret += tree1[i];
  return ret;
}
long long tree2[MAXN + 10];
void update2(int i, long long v) {
  for (; i <= N; i += (i & -i)) tree2[i] += v;
}
long long query2(int i) {
  long long ret = 0;
  for (; i > 0; i -= (i & -i)) ret += tree2[i];
  return ret;
}
int main() {
  int i, j;
  scanf("%d", &N);
  for (i = 1; i <= N; i++) scanf("%lld", &A[i].first);
  for (i = 1; i <= N; i++)
    scanf("%lld", &A[i].second), comp.push_back(A[i].second);
  sort(A + 1, A + N + 1);
  sort(comp.begin(), comp.end());
  comp.erase(unique(comp.begin(), comp.end()), comp.end());
  for (i = 1; i <= N; i++)
    A[i].second =
        lower_bound(comp.begin(), comp.end(), A[i].second) - comp.begin() + 1;
  for (i = 1; i <= N; i++) {
    ans += query2(A[i].second) * A[i].first - query1(A[i].second);
    update1(A[i].second, A[i].first);
    update2(A[i].second, 1);
  }
  printf("%lld\n", ans);
}

### 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;
const int MAXN = 2e5;
int N;
pair<long long, long long> A[MAXN + 10];
long long ans;
vector<long long> comp;
long long tree1[MAXN + 10];
void update1(int i, long long v) {
  for (; i <= N; i += (i & -i)) tree1[i] += v;
}
long long query1(int i) {
  long long ret = 0;
  for (; i > 0; i -= (i & -i)) ret += tree1[i];
  return ret;
}
long long tree2[MAXN + 10];
void update2(int i, long long v) {
  for (; i <= N; i += (i & -i)) tree2[i] += v;
}
long long query2(int i) {
  long long ret = 0;
  for (; i > 0; i -= (i & -i)) ret += tree2[i];
  return ret;
}
int main() {
  int i, j;
  scanf("%d", &N);
  for (i = 1; i <= N; i++) scanf("%lld", &A[i].first);
  for (i = 1; i <= N; i++)
    scanf("%lld", &A[i].second), comp.push_back(A[i].second);
  sort(A + 1, A + N + 1);
  sort(comp.begin(), comp.end());
  comp.erase(unique(comp.begin(), comp.end()), comp.end());
  for (i = 1; i <= N; i++)
    A[i].second =
        lower_bound(comp.begin(), comp.end(), A[i].second) - comp.begin() + 1;
  for (i = 1; i <= N; i++) {
    ans += query2(A[i].second) * A[i].first - query1(A[i].second);
    update1(A[i].second, A[i].first);
    update2(A[i].second, 1);
  }
  printf("%lld\n", ans);
}

```