#include <bits/stdc++.h>
using namespace std;
void dout() { cerr << '\n'; }
template <typename Head, typename... Tail>
void dout(Head H, Tail... T) {
  cerr << " " << H;
  dout(T...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5 + 7;
int n, m, cnt[N];
long long sum[N];
pair<int, int> p[N];
void upd(int x, int y) {
  for (; x <= m; x |= (x + 1)) {
    cnt[x]++;
    sum[x] += y;
  }
}
int getnum(int x) {
  int res = 0;
  for (; x >= 0; x = (x & (x + 1)) - 1) {
    res += cnt[x];
  }
  return res;
}
long long getsum(int x) {
  long long res = 0;
  for (; x >= 0; x = (x & (x + 1)) - 1) {
    res += sum[x];
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  cin >> n;
  vector<int> vec, v;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].first;
  }
  for (int i = 1; i <= n; i++) {
    cin >> p[i].second;
    vec.push_back(p[i].second);
  }
  sort((vec).begin(), (vec).end());
  map<int, int> ind;
  for (int i = 1; i <= n; i++) {
    if (!ind.count(vec[i - 1])) {
      ind[vec[i - 1]] = ++m;
    }
  }
  for (int i = 1; i <= n; i++) {
    p[i].second = ind[p[i].second];
  }
  sort(p + 1, p + n + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += (long long)getnum(p[i].second) * p[i].first - getsum(p[i].second);
    upd(p[i].second, p[i].first);
  }
  cout << ans;
}

### Prompt

Create a solution in CPP for 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;
void dout() { cerr << '\n'; }
template <typename Head, typename... Tail>
void dout(Head H, Tail... T) {
  cerr << " " << H;
  dout(T...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5 + 7;
int n, m, cnt[N];
long long sum[N];
pair<int, int> p[N];
void upd(int x, int y) {
  for (; x <= m; x |= (x + 1)) {
    cnt[x]++;
    sum[x] += y;
  }
}
int getnum(int x) {
  int res = 0;
  for (; x >= 0; x = (x & (x + 1)) - 1) {
    res += cnt[x];
  }
  return res;
}
long long getsum(int x) {
  long long res = 0;
  for (; x >= 0; x = (x & (x + 1)) - 1) {
    res += sum[x];
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  cin >> n;
  vector<int> vec, v;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].first;
  }
  for (int i = 1; i <= n; i++) {
    cin >> p[i].second;
    vec.push_back(p[i].second);
  }
  sort((vec).begin(), (vec).end());
  map<int, int> ind;
  for (int i = 1; i <= n; i++) {
    if (!ind.count(vec[i - 1])) {
      ind[vec[i - 1]] = ++m;
    }
  }
  for (int i = 1; i <= n; i++) {
    p[i].second = ind[p[i].second];
  }
  sort(p + 1, p + n + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += (long long)getnum(p[i].second) * p[i].first - getsum(p[i].second);
    upd(p[i].second, p[i].first);
  }
  cout << ans;
}

```