#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 7;
int n;
vector<pair<int, int>> p;
vector<int> vv1;
long long get(vector<long long> &f, int pos) {
  long long res = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) {
    res += f[pos];
  }
  return res;
}
void upd(vector<long long> &f, int pos, int val) {
  for (; pos < f.size(); pos |= pos + 1) f[pos] += val;
}
int main() {
  scanf("%d", &n);
  p.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &p[i].first);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i].second);
    vv1.push_back(p[i].second);
  }
  sort(vv1.begin(), vv1.end());
  vv1.resize(unique(vv1.begin(), vv1.end()) - vv1.begin());
  sort(p.begin(), p.end());
  long long ans = 0;
  vector<long long> tmpv1(vv1.size()), tmpv2(vv1.size());
  for (int i = 0; i < n; i++) {
    int pos = lower_bound(vv1.begin(), vv1.end(), p[i].second) - vv1.begin();
    ans += get(tmpv1, pos) * 1ll * p[i].first - get(tmpv2, pos);
    upd(tmpv1, pos, 1);
    upd(tmpv2, pos, p[i].first);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 N = 1e5 + 7;
int n;
vector<pair<int, int>> p;
vector<int> vv1;
long long get(vector<long long> &f, int pos) {
  long long res = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) {
    res += f[pos];
  }
  return res;
}
void upd(vector<long long> &f, int pos, int val) {
  for (; pos < f.size(); pos |= pos + 1) f[pos] += val;
}
int main() {
  scanf("%d", &n);
  p.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &p[i].first);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i].second);
    vv1.push_back(p[i].second);
  }
  sort(vv1.begin(), vv1.end());
  vv1.resize(unique(vv1.begin(), vv1.end()) - vv1.begin());
  sort(p.begin(), p.end());
  long long ans = 0;
  vector<long long> tmpv1(vv1.size()), tmpv2(vv1.size());
  for (int i = 0; i < n; i++) {
    int pos = lower_bound(vv1.begin(), vv1.end(), p[i].second) - vv1.begin();
    ans += get(tmpv1, pos) * 1ll * p[i].first - get(tmpv2, pos);
    upd(tmpv1, pos, 1);
    upd(tmpv2, pos, p[i].first);
  }
  printf("%lld\n", ans);
  return 0;
}

```