#include <bits/stdc++.h>
using namespace std;
template <class T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
void setmax(T& a, T b) {
  a = max(a, b);
};
template <typename T>
void setmin(T& a, T b) {
  a = min(a, b);
};
template <typename T>
bool in(T v, T lo, T hi) {
  return lo <= v && v <= hi;
};
const int N = 2e5 + 10;
int n;
long long fena[N], fenb[N];
void add(long long* first, int i, long long delta) {
  for (int j = i; j < N; j += j & -j) {
    first[j] += delta;
  }
}
long long sum(long long* first, int i) {
  long long res = 0;
  for (int j = i; j; j -= j & -j) {
    res += first[j];
  }
  return res;
}
struct FastCoordinateCompression {
  vector<int> a;
  void add(int x) { a.push_back(x); }
  void compress() {
    sort(a.begin(), a.end());
    a.resize(unique(a.begin(), a.end()) - a.begin());
  }
  int operator[](int x) {
    return lower_bound(a.begin(), a.end(), x) - a.begin() + 1;
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<pair<int, int>> xv(n);
  for (auto& [x, v] : xv) {
    cin >> x;
  }
  for (auto& [x, v] : xv) {
    cin >> v;
  }
  FastCoordinateCompression compr;
  for (auto& [x, v] : xv) {
    compr.add(x);
  }
  compr.compress();
  sort(xv.begin(), xv.end(), [&](pair<int, int>& a, pair<int, int>& b) {
    if (a.second != b.second) return a.second < b.second;
    return a.first < b.first;
  });
  long long ans = 0;
  for (auto& [x, v] : xv) {
    long long num = sum(fena, compr[x]);
    long long tot = sum(fenb, compr[x]);
    ans += x * num - tot;
    add(fena, compr[x], 1);
    add(fenb, compr[x], x);
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
template <class T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
void setmax(T& a, T b) {
  a = max(a, b);
};
template <typename T>
void setmin(T& a, T b) {
  a = min(a, b);
};
template <typename T>
bool in(T v, T lo, T hi) {
  return lo <= v && v <= hi;
};
const int N = 2e5 + 10;
int n;
long long fena[N], fenb[N];
void add(long long* first, int i, long long delta) {
  for (int j = i; j < N; j += j & -j) {
    first[j] += delta;
  }
}
long long sum(long long* first, int i) {
  long long res = 0;
  for (int j = i; j; j -= j & -j) {
    res += first[j];
  }
  return res;
}
struct FastCoordinateCompression {
  vector<int> a;
  void add(int x) { a.push_back(x); }
  void compress() {
    sort(a.begin(), a.end());
    a.resize(unique(a.begin(), a.end()) - a.begin());
  }
  int operator[](int x) {
    return lower_bound(a.begin(), a.end(), x) - a.begin() + 1;
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<pair<int, int>> xv(n);
  for (auto& [x, v] : xv) {
    cin >> x;
  }
  for (auto& [x, v] : xv) {
    cin >> v;
  }
  FastCoordinateCompression compr;
  for (auto& [x, v] : xv) {
    compr.add(x);
  }
  compr.compress();
  sort(xv.begin(), xv.end(), [&](pair<int, int>& a, pair<int, int>& b) {
    if (a.second != b.second) return a.second < b.second;
    return a.first < b.first;
  });
  long long ans = 0;
  for (auto& [x, v] : xv) {
    long long num = sum(fena, compr[x]);
    long long tot = sum(fenb, compr[x]);
    ans += x * num - tot;
    add(fena, compr[x], 1);
    add(fenb, compr[x], x);
  }
  cout << ans << '\n';
  return 0;
}

```