#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5;
int m;
long long val[maxn + 11];
int cnt[maxn + 11];
pair<int, int> p[maxn + 11];
vector<int> v;
int lowbit(int x) { return x & (-x); }
void add(int x, int o) {
  for (; x <= m; x += lowbit(x)) {
    cnt[x] += 1;
    val[x] += o;
  }
}
pair<int, long long> query(int x) {
  int resc = 0;
  long long resv = 0;
  for (; x; x -= lowbit(x)) {
    resc += cnt[x];
    resv += val[x];
  }
  return make_pair(resc, resv);
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i].first;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].second;
    v.emplace_back(p[i].second);
  }
  sort(p + 1, p + 1 + n);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  m = v.size();
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int pos = lower_bound(v.begin(), v.end(), p[i].second) - v.begin() + 1;
    pair<int, long long> pi = query(pos);
    ans += 1ll * pi.first * p[i].first - pi.second;
    add(pos, p[i].first);
  }
  cout << ans << endl;
}

### Prompt

In Cpp, your task is to solve 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 int maxn = 2e5;
int m;
long long val[maxn + 11];
int cnt[maxn + 11];
pair<int, int> p[maxn + 11];
vector<int> v;
int lowbit(int x) { return x & (-x); }
void add(int x, int o) {
  for (; x <= m; x += lowbit(x)) {
    cnt[x] += 1;
    val[x] += o;
  }
}
pair<int, long long> query(int x) {
  int resc = 0;
  long long resv = 0;
  for (; x; x -= lowbit(x)) {
    resc += cnt[x];
    resv += val[x];
  }
  return make_pair(resc, resv);
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i].first;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].second;
    v.emplace_back(p[i].second);
  }
  sort(p + 1, p + 1 + n);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  m = v.size();
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int pos = lower_bound(v.begin(), v.end(), p[i].second) - v.begin() + 1;
    pair<int, long long> pi = query(pos);
    ans += 1ll * pi.first * p[i].first - pi.second;
    add(pos, p[i].first);
  }
  cout << ans << endl;
}

```