#include <bits/stdc++.h>
long double pi = acos(-1.0);
const int mod = 1e9 + 7;
using namespace std;
const int N = 2e5 + 5;
vector<long long> F(N), F2(N);
long long sum(int r, vector<long long> &t) {
  long long result = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) result += t[r];
  return result;
}
void inc(int i, long long delta, vector<long long> &t) {
  for (; i < N; i = (i | (i + 1))) t[i] += delta;
}
long long sum(int l, int r, vector<long long> &t) {
  return sum(r, t) - sum(l - 1, t);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  cout.setf(ios::fixed);
  int n;
  cin >> n;
  vector<pair<int, int> > pts(n);
  for (int(i) = 0; (i) < n; (i)++) {
    cin >> pts[i].first;
  }
  vector<int> vv;
  for (int(i) = 0; (i) < n; (i)++) {
    cin >> pts[i].second;
    vv.push_back(pts[i].second);
  }
  sort((vv).begin(), (vv).end());
  vv.resize(unique((vv).begin(), (vv).end()) - vv.begin());
  for (int(i) = 0; (i) < n; (i)++) {
    pts[i].second =
        lower_bound((vv).begin(), (vv).end(), pts[i].second) - vv.begin();
  }
  sort((pts).begin(), (pts).end());
  for (int(i) = 0; (i) < n; (i)++) {
    inc(pts[i].second, pts[i].first, F);
    inc(pts[i].second, 1, F2);
  }
  long long ans = 0;
  for (int(i) = 0; (i) < n; (i)++) {
    int v = pts[i].second;
    inc(v, -pts[i].first, F);
    inc(v, -1, F2);
    long long sum1 = sum(v, N - 1, F);
    long long cnt = sum(v, N - 1, F2);
    ans += sum1 - cnt * (long long)pts[i].first;
  }
  cout << ans;
  return 0;
}

### 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>
long double pi = acos(-1.0);
const int mod = 1e9 + 7;
using namespace std;
const int N = 2e5 + 5;
vector<long long> F(N), F2(N);
long long sum(int r, vector<long long> &t) {
  long long result = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) result += t[r];
  return result;
}
void inc(int i, long long delta, vector<long long> &t) {
  for (; i < N; i = (i | (i + 1))) t[i] += delta;
}
long long sum(int l, int r, vector<long long> &t) {
  return sum(r, t) - sum(l - 1, t);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  cout.setf(ios::fixed);
  int n;
  cin >> n;
  vector<pair<int, int> > pts(n);
  for (int(i) = 0; (i) < n; (i)++) {
    cin >> pts[i].first;
  }
  vector<int> vv;
  for (int(i) = 0; (i) < n; (i)++) {
    cin >> pts[i].second;
    vv.push_back(pts[i].second);
  }
  sort((vv).begin(), (vv).end());
  vv.resize(unique((vv).begin(), (vv).end()) - vv.begin());
  for (int(i) = 0; (i) < n; (i)++) {
    pts[i].second =
        lower_bound((vv).begin(), (vv).end(), pts[i].second) - vv.begin();
  }
  sort((pts).begin(), (pts).end());
  for (int(i) = 0; (i) < n; (i)++) {
    inc(pts[i].second, pts[i].first, F);
    inc(pts[i].second, 1, F2);
  }
  long long ans = 0;
  for (int(i) = 0; (i) < n; (i)++) {
    int v = pts[i].second;
    inc(v, -pts[i].first, F);
    inc(v, -1, F2);
    long long sum1 = sum(v, N - 1, F);
    long long cnt = sum(v, N - 1, F2);
    ans += sum1 - cnt * (long long)pts[i].first;
  }
  cout << ans;
  return 0;
}

```