#include <bits/stdc++.h>
using namespace std;
inline int lowbit(int x) { return x & (-x); }
long long getsum(vector<long long> &V, int pos) {
  long long ret = 0;
  for (; pos; pos -= lowbit(pos)) ret += V[pos];
  return ret;
}
void add(vector<long long> &V, int pos, int k) {
  for (; pos < (int)V.size(); pos += lowbit(pos)) V[pos] += k;
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int>> p(n);
  for (auto &pnt : p) {
    int x;
    scanf("%d", &x);
    pnt.first = x;
  }
  for (auto &pnt : p) {
    int v;
    scanf("%d", &v);
    pnt.second = v;
  }
  sort(p.begin(), p.end());
  vector<int> vs{(int)(-1e9)};
  for (auto pnt : p) vs.push_back(pnt.second);
  sort(vs.begin(), vs.end());
  vs.resize(unique(vs.begin(), vs.end()) - vs.begin());
  long long ans = 0;
  vector<long long> cnt(vs.size()), xs(vs.size());
  for (auto pnt : p) {
    int pos = lower_bound(vs.begin(), vs.end(), pnt.second) - vs.begin();
    ans += getsum(cnt, pos) * pnt.first - getsum(xs, pos);
    add(cnt, pos, 1), add(xs, pos, pnt.first);
  }
  printf("%lld\n", ans);
}

### 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;
inline int lowbit(int x) { return x & (-x); }
long long getsum(vector<long long> &V, int pos) {
  long long ret = 0;
  for (; pos; pos -= lowbit(pos)) ret += V[pos];
  return ret;
}
void add(vector<long long> &V, int pos, int k) {
  for (; pos < (int)V.size(); pos += lowbit(pos)) V[pos] += k;
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int>> p(n);
  for (auto &pnt : p) {
    int x;
    scanf("%d", &x);
    pnt.first = x;
  }
  for (auto &pnt : p) {
    int v;
    scanf("%d", &v);
    pnt.second = v;
  }
  sort(p.begin(), p.end());
  vector<int> vs{(int)(-1e9)};
  for (auto pnt : p) vs.push_back(pnt.second);
  sort(vs.begin(), vs.end());
  vs.resize(unique(vs.begin(), vs.end()) - vs.begin());
  long long ans = 0;
  vector<long long> cnt(vs.size()), xs(vs.size());
  for (auto pnt : p) {
    int pos = lower_bound(vs.begin(), vs.end(), pnt.second) - vs.begin();
    ans += getsum(cnt, pos) * pnt.first - getsum(xs, pos);
    add(cnt, pos, 1), add(xs, pos, pnt.first);
  }
  printf("%lld\n", ans);
}

```