#include <bits/stdc++.h>
using namespace std;
struct FenwickTree {
  int n;
  vector<long long> num;
  FenwickTree() : n(0) {}
  FenwickTree(int _n) {
    n = _n;
    num.assign(n, 0);
  }
  void add(int i, int val) {
    for (; i < n; i |= i + 1) num[i] += val;
  }
  long long sum(int i) {
    long long ret = 0;
    for (; i >= 0; i = (i & (i + 1)) - 1) ret += num[i];
    return ret;
  }
};
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, int>> point(n);
  vector<int> speed;
  for (int i = 0; i < n; i++) scanf("%d", &point[i].first);
  for (int i = 0; i < n; i++) {
    scanf("%d", &point[i].second);
    speed.push_back(point[i].second);
  }
  sort(speed.begin(), speed.end());
  speed.erase(unique(speed.begin(), speed.end()), speed.end());
  sort(point.begin(), point.end());
  long long ans = 0;
  FenwickTree cnt(n), sumx(n);
  for (auto i : point) {
    i.second =
        lower_bound(speed.begin(), speed.end(), i.second) - speed.begin();
    ans += cnt.sum(i.second) * i.first - sumx.sum(i.second);
    cnt.add(i.second, 1);
    sumx.add(i.second, i.first);
  }
  printf("%lld\n", ans);
  return 0;
}

### 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;
struct FenwickTree {
  int n;
  vector<long long> num;
  FenwickTree() : n(0) {}
  FenwickTree(int _n) {
    n = _n;
    num.assign(n, 0);
  }
  void add(int i, int val) {
    for (; i < n; i |= i + 1) num[i] += val;
  }
  long long sum(int i) {
    long long ret = 0;
    for (; i >= 0; i = (i & (i + 1)) - 1) ret += num[i];
    return ret;
  }
};
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, int>> point(n);
  vector<int> speed;
  for (int i = 0; i < n; i++) scanf("%d", &point[i].first);
  for (int i = 0; i < n; i++) {
    scanf("%d", &point[i].second);
    speed.push_back(point[i].second);
  }
  sort(speed.begin(), speed.end());
  speed.erase(unique(speed.begin(), speed.end()), speed.end());
  sort(point.begin(), point.end());
  long long ans = 0;
  FenwickTree cnt(n), sumx(n);
  for (auto i : point) {
    i.second =
        lower_bound(speed.begin(), speed.end(), i.second) - speed.begin();
    ans += cnt.sum(i.second) * i.first - sumx.sum(i.second);
    cnt.add(i.second, 1);
    sumx.add(i.second, i.first);
  }
  printf("%lld\n", ans);
  return 0;
}

```