#include <bits/stdc++.h>
using namespace std;
struct Pt {
  int x, v;
  bool operator<(const Pt &a) const { return x < a.x; }
};
struct BIT {
  vector<long long> bit, cnt;
  int n;
  BIT(int sz) {
    bit.resize(sz + 1);
    cnt.resize(sz + 1);
    n = sz;
  }
  void add(int x, long long v) {
    while (x <= n) {
      cnt[x]++;
      bit[x] += v;
      x += x & (-x);
    }
  }
  pair<long long, long long> ask(int x) {
    pair<long long, long long> ret(0, 0);
    while (x) {
      ret.first += bit[x];
      ret.second += cnt[x];
      x -= x & (-x);
    }
    return ret;
  }
};
vector<int> V;
Pt arr[200005];
int n;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i].x);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i].v);
    V.push_back(arr[i].v);
  }
  long long ans = 0;
  sort(arr, arr + n);
  sort(begin(V), end(V));
  V.erase(unique(begin(V), end(V)), end(V));
  BIT loli(V.size());
  for (int i = 0; i < n; i++) {
    int p = lower_bound(begin(V), end(V), arr[i].v) - begin(V) + 1;
    auto val = loli.ask(p);
    ans = ans + arr[i].x * val.second - val.first;
    loli.add(p, arr[i].x);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your task is to create 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 Pt {
  int x, v;
  bool operator<(const Pt &a) const { return x < a.x; }
};
struct BIT {
  vector<long long> bit, cnt;
  int n;
  BIT(int sz) {
    bit.resize(sz + 1);
    cnt.resize(sz + 1);
    n = sz;
  }
  void add(int x, long long v) {
    while (x <= n) {
      cnt[x]++;
      bit[x] += v;
      x += x & (-x);
    }
  }
  pair<long long, long long> ask(int x) {
    pair<long long, long long> ret(0, 0);
    while (x) {
      ret.first += bit[x];
      ret.second += cnt[x];
      x -= x & (-x);
    }
    return ret;
  }
};
vector<int> V;
Pt arr[200005];
int n;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i].x);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i].v);
    V.push_back(arr[i].v);
  }
  long long ans = 0;
  sort(arr, arr + n);
  sort(begin(V), end(V));
  V.erase(unique(begin(V), end(V)), end(V));
  BIT loli(V.size());
  for (int i = 0; i < n; i++) {
    int p = lower_bound(begin(V), end(V), arr[i].v) - begin(V) + 1;
    auto val = loli.ask(p);
    ans = ans + arr[i].x * val.second - val.first;
    loli.add(p, arr[i].x);
  }
  printf("%lld\n", ans);
  return 0;
}

```