#include <bits/stdc++.h>
using namespace std;
const long long mx = 2 * 1e5 + 5;
const long long mod = 1e9 + 7;
long long sum[mx];
long long num[mx];
void update(int pos, int value) {
  while (pos <= mx) {
    sum[pos] += value;
    num[pos]++;
    pos += (pos & -pos);
  }
}
pair<int, long long> query(int pos) {
  long long summ = 0;
  long long numm = 0;
  while (pos >= 1) {
    summ += sum[pos];
    numm += num[pos];
    pos -= (pos & -pos);
  }
  return {numm, summ};
}
int main() {
  int n;
  cin >> n;
  pair<long long, long long> p[n];
  map<long long, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> p[i].first;
  }
  for (int i = 0; i < n; i++) {
    cin >> p[i].second;
    mp[p[i].second] = i;
  }
  int cur = 1LL;
  for (auto& i : mp) {
    i.second = cur;
    cur++;
  }
  for (int i = 0; i < n; i++) {
    p[i].second = mp[p[i].second];
  }
  sort(p, p + n);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long myv = p[i].second;
    long long myx = p[i].first;
    pair<long long, long long> cnt = query(myv);
    ans += (1LL * myx * cnt.first);
    ans -= cnt.second;
    update(myv, myx);
  }
  cout << ans;
}

### Prompt

Generate 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 long long mx = 2 * 1e5 + 5;
const long long mod = 1e9 + 7;
long long sum[mx];
long long num[mx];
void update(int pos, int value) {
  while (pos <= mx) {
    sum[pos] += value;
    num[pos]++;
    pos += (pos & -pos);
  }
}
pair<int, long long> query(int pos) {
  long long summ = 0;
  long long numm = 0;
  while (pos >= 1) {
    summ += sum[pos];
    numm += num[pos];
    pos -= (pos & -pos);
  }
  return {numm, summ};
}
int main() {
  int n;
  cin >> n;
  pair<long long, long long> p[n];
  map<long long, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> p[i].first;
  }
  for (int i = 0; i < n; i++) {
    cin >> p[i].second;
    mp[p[i].second] = i;
  }
  int cur = 1LL;
  for (auto& i : mp) {
    i.second = cur;
    cur++;
  }
  for (int i = 0; i < n; i++) {
    p[i].second = mp[p[i].second];
  }
  sort(p, p + n);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long myv = p[i].second;
    long long myx = p[i].first;
    pair<long long, long long> cnt = query(myv);
    ans += (1LL * myx * cnt.first);
    ans -= cnt.second;
    update(myv, myx);
  }
  cout << ans;
}

```