#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const int MAX = 2e5 + 10;
vector<pair<int, int>> v(MAX);
vector<int> spdv(MAX), compress(MAX);
long long n, bit[2][MAX];
void poe(int a, long long x, long long p) {
  for (; p <= n; p += p & -p) bit[a][p] += x;
}
long long pref(int a, long long p) {
  long long ret = 0;
  for (; p; p -= p & -p) ret += bit[a][p];
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> v[i].first;
    v[i].second = i;
  }
  for (int i = 0; i < n; i++) {
    cin >> spdv[i];
    compress[i] = spdv[i];
  }
  sort(v.begin(), v.begin() + n);
  sort(compress.begin(), compress.begin() + n);
  map<int, int> newidx;
  for (int i = 0; i < n; i++) newidx[compress[i]] = i;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int pos = newidx[spdv[v[i].second]] + 1;
    ans += v[i].first * pref(0, pos) - pref(1, pos);
    poe(0, 1, pos);
    poe(1, v[i].first, pos);
  }
  cout << ans << '\n';
  return 0;
}

### 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 int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const int MAX = 2e5 + 10;
vector<pair<int, int>> v(MAX);
vector<int> spdv(MAX), compress(MAX);
long long n, bit[2][MAX];
void poe(int a, long long x, long long p) {
  for (; p <= n; p += p & -p) bit[a][p] += x;
}
long long pref(int a, long long p) {
  long long ret = 0;
  for (; p; p -= p & -p) ret += bit[a][p];
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> v[i].first;
    v[i].second = i;
  }
  for (int i = 0; i < n; i++) {
    cin >> spdv[i];
    compress[i] = spdv[i];
  }
  sort(v.begin(), v.begin() + n);
  sort(compress.begin(), compress.begin() + n);
  map<int, int> newidx;
  for (int i = 0; i < n; i++) newidx[compress[i]] = i;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int pos = newidx[spdv[v[i].second]] + 1;
    ans += v[i].first * pref(0, pos) - pref(1, pos);
    poe(0, 1, pos);
    poe(1, v[i].first, pos);
  }
  cout << ans << '\n';
  return 0;
}

```