#include <bits/stdc++.h>
using namespace std;
long long ans, n, a1[202020], a2[202020], od[202020];
pair<long long, long long> p[202020];
map<long long, long long> mp;
void gx1(long long x, long long v) {
  for (long long i = x; i < 200200; i += i & -i) {
    a1[i] += v;
  }
}
void gx2(long long x, long long v) {
  for (long long i = x; i < 200200; i += i & -i) {
    a2[i] += v;
  }
}
long long gt1(long long p) {
  long long rt = 0;
  for (long long i = p; i > 0; i -= i & -i) {
    rt += a1[i];
  }
  return rt;
}
long long gt2(long long p) {
  long long rt = 0;
  for (long long i = p; i > 0; i -= i & -i) {
    rt += a2[i];
  }
  return rt;
}
signed main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> p[i].first;
  }
  for (long long i = 1; i <= n; i++) {
    cin >> p[i].second;
    od[i] = p[i].second;
  }
  sort(od + 1, od + n + 1);
  sort(p + 1, p + n + 1);
  long long pp = 1;
  for (long long i = 1; i <= n; i++) {
    if (!mp[od[i]]) {
      mp[od[i]] = pp;
      pp++;
    }
  }
  for (long long i = 1; i <= n; i++) {
    ans += p[i].first * gt1(mp[p[i].second]) - gt2(mp[p[i].second]);
    gx1(mp[p[i].second], 1ll);
    gx2(mp[p[i].second], p[i].first);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
long long ans, n, a1[202020], a2[202020], od[202020];
pair<long long, long long> p[202020];
map<long long, long long> mp;
void gx1(long long x, long long v) {
  for (long long i = x; i < 200200; i += i & -i) {
    a1[i] += v;
  }
}
void gx2(long long x, long long v) {
  for (long long i = x; i < 200200; i += i & -i) {
    a2[i] += v;
  }
}
long long gt1(long long p) {
  long long rt = 0;
  for (long long i = p; i > 0; i -= i & -i) {
    rt += a1[i];
  }
  return rt;
}
long long gt2(long long p) {
  long long rt = 0;
  for (long long i = p; i > 0; i -= i & -i) {
    rt += a2[i];
  }
  return rt;
}
signed main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> p[i].first;
  }
  for (long long i = 1; i <= n; i++) {
    cin >> p[i].second;
    od[i] = p[i].second;
  }
  sort(od + 1, od + n + 1);
  sort(p + 1, p + n + 1);
  long long pp = 1;
  for (long long i = 1; i <= n; i++) {
    if (!mp[od[i]]) {
      mp[od[i]] = pp;
      pp++;
    }
  }
  for (long long i = 1; i <= n; i++) {
    ans += p[i].first * gt1(mp[p[i].second]) - gt2(mp[p[i].second]);
    gx1(mp[p[i].second], 1ll);
    gx2(mp[p[i].second], p[i].first);
  }
  cout << ans << endl;
  return 0;
}

```