#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200200;
int n;
long long ans, bit[2][MAXN];
pair<int, int> a[MAXN];
void compress() {
  map<int, int> m;
  for (int i = 1; i <= n; ++i) m[a[i].second];
  int cnt = 1;
  for (auto& [key, val] : m) val = cnt++;
  for (int i = 1; i <= n; ++i) a[i].second = m[a[i].second];
}
void add(int b, int pos, int val) {
  for (; 0 < pos && pos < MAXN; pos += pos & -pos) bit[b][pos] += val;
}
long long get(int b, int pos) {
  long long ret = 0;
  for (; 0 < pos && pos < MAXN; pos -= pos & -pos) ret += bit[b][pos];
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].first);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].second);
  sort(a + 1, a + n + 1);
  compress();
  for (int i = 1; i <= n; ++i) {
    auto [x, v] = a[i];
    ans += 1LL * x * get(0, v) - get(1, v);
    add(1, v, x);
    add(0, v, 1);
  }
  printf("%lld\n", ans);
}

### Prompt

Your challenge is to write 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 int MAXN = 200200;
int n;
long long ans, bit[2][MAXN];
pair<int, int> a[MAXN];
void compress() {
  map<int, int> m;
  for (int i = 1; i <= n; ++i) m[a[i].second];
  int cnt = 1;
  for (auto& [key, val] : m) val = cnt++;
  for (int i = 1; i <= n; ++i) a[i].second = m[a[i].second];
}
void add(int b, int pos, int val) {
  for (; 0 < pos && pos < MAXN; pos += pos & -pos) bit[b][pos] += val;
}
long long get(int b, int pos) {
  long long ret = 0;
  for (; 0 < pos && pos < MAXN; pos -= pos & -pos) ret += bit[b][pos];
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].first);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].second);
  sort(a + 1, a + n + 1);
  compress();
  for (int i = 1; i <= n; ++i) {
    auto [x, v] = a[i];
    ans += 1LL * x * get(0, v) - get(1, v);
    add(1, v, x);
    add(0, v, 1);
  }
  printf("%lld\n", ans);
}

```