#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5;
long long Bit[N][2], n;
long long get(long long i, long long k) {
  long long ans = 0;
  for (i; i > 0; i -= i & (-i)) {
    ans += Bit[i][k];
  }
  return ans;
}
void update(long long i, long long val, long long k) {
  for (i; i <= n; i += i & (-i)) {
    Bit[i][k] += val;
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long k, i, t, j, m, x, y, z, c = 0, ans = 0;
  cin >> n;
  vector<pair<long long, long long> > v(n), v1(n);
  for (i = 0; i < n; i++) cin >> v[i].first;
  for (i = 0; i < n; i++) cin >> v[i].second;
  sort((v).begin(), (v).end());
  vector<long long> v0(n);
  map<long long, long long> mp;
  c = 1;
  for (i = 0; i < n; i++) {
    v1[i].first = v[i].second;
    v1[i].second = i;
  }
  sort((v1).begin(), (v1).end());
  for (i = 0; i < n; i++) {
    if (mp[v1[i].first] == 0) {
      mp[v1[i].first] = c;
      c++;
    }
    v0[v1[i].second] = mp[v1[i].first];
  }
  for (i = 0; i < n; i++) {
    x = get(v0[i], 0);
    y = get(v0[i], 1);
    ans += x * v[i].first - y;
    update(v0[i], 1, 0);
    update(v0[i], v[i].first, 1);
  }
  cout << ans;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 long long N = 2e5 + 5;
long long Bit[N][2], n;
long long get(long long i, long long k) {
  long long ans = 0;
  for (i; i > 0; i -= i & (-i)) {
    ans += Bit[i][k];
  }
  return ans;
}
void update(long long i, long long val, long long k) {
  for (i; i <= n; i += i & (-i)) {
    Bit[i][k] += val;
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long k, i, t, j, m, x, y, z, c = 0, ans = 0;
  cin >> n;
  vector<pair<long long, long long> > v(n), v1(n);
  for (i = 0; i < n; i++) cin >> v[i].first;
  for (i = 0; i < n; i++) cin >> v[i].second;
  sort((v).begin(), (v).end());
  vector<long long> v0(n);
  map<long long, long long> mp;
  c = 1;
  for (i = 0; i < n; i++) {
    v1[i].first = v[i].second;
    v1[i].second = i;
  }
  sort((v1).begin(), (v1).end());
  for (i = 0; i < n; i++) {
    if (mp[v1[i].first] == 0) {
      mp[v1[i].first] = c;
      c++;
    }
    v0[v1[i].second] = mp[v1[i].first];
  }
  for (i = 0; i < n; i++) {
    x = get(v0[i], 0);
    y = get(v0[i], 1);
    ans += x * v[i].first - y;
    update(v0[i], 1, 0);
    update(v0[i], v[i].first, 1);
  }
  cout << ans;
}

```