#include <bits/stdc++.h>
using namespace std;
long long int n, bit1[200050], bit2[200050];
vector<pair<long long int, long long int> > arr(200050);
map<long long int, long long int> mp;
long long int get1(int idx) {
  long long int ans = 0;
  while (idx > 0) {
    ans += bit1[idx];
    idx -= (idx & -idx);
  }
  return ans;
}
long long int get2(int idx) {
  long long int ans = 0;
  while (idx > 0) {
    ans += bit2[idx];
    idx -= (idx & -idx);
  }
  return ans;
}
void update1(int idx, int inc) {
  while (idx <= n) {
    bit1[idx] += inc;
    idx += (idx & -idx);
  }
}
void update2(int idx, int inc) {
  while (idx <= n) {
    bit2[idx] += inc;
    idx += (idx & -idx);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  arr.resize(n);
  for (int i = 0; i < n; i++) cin >> arr[i].first;
  for (int i = 0; i < n; i++) cin >> arr[i].second;
  sort(arr.begin(), arr.end());
  vector<long long int> v;
  for (auto a : arr) v.push_back(a.second);
  sort(v.begin(), v.end());
  int p = 1;
  for (auto x : v) mp[x] = p++;
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    long long int cnt = get1(mp[arr[i].second]);
    long long int sum = get2(mp[arr[i].second]);
    ans += cnt * arr[i].first - sum;
    update1(mp[arr[i].second], 1);
    update2(mp[arr[i].second], arr[i].first);
  }
  cout << 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;
long long int n, bit1[200050], bit2[200050];
vector<pair<long long int, long long int> > arr(200050);
map<long long int, long long int> mp;
long long int get1(int idx) {
  long long int ans = 0;
  while (idx > 0) {
    ans += bit1[idx];
    idx -= (idx & -idx);
  }
  return ans;
}
long long int get2(int idx) {
  long long int ans = 0;
  while (idx > 0) {
    ans += bit2[idx];
    idx -= (idx & -idx);
  }
  return ans;
}
void update1(int idx, int inc) {
  while (idx <= n) {
    bit1[idx] += inc;
    idx += (idx & -idx);
  }
}
void update2(int idx, int inc) {
  while (idx <= n) {
    bit2[idx] += inc;
    idx += (idx & -idx);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  arr.resize(n);
  for (int i = 0; i < n; i++) cin >> arr[i].first;
  for (int i = 0; i < n; i++) cin >> arr[i].second;
  sort(arr.begin(), arr.end());
  vector<long long int> v;
  for (auto a : arr) v.push_back(a.second);
  sort(v.begin(), v.end());
  int p = 1;
  for (auto x : v) mp[x] = p++;
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    long long int cnt = get1(mp[arr[i].second]);
    long long int sum = get2(mp[arr[i].second]);
    ans += cnt * arr[i].first - sum;
    update1(mp[arr[i].second], 1);
    update2(mp[arr[i].second], arr[i].first);
  }
  cout << ans;
  return 0;
}

```