#include <bits/stdc++.h>
using namespace std;
template <typename T>
struct BIT {
  int n;
  vector<T> bit;
  BIT(int n_ = 0) : n(n_), bit(n_ + 1) {}
  T sum(int i) {
    T ans = 0;
    for (; i > 0; i -= i & -i) ans += bit[i];
    return ans;
  }
  void add(int i, T a) {
    if (i == 0) return;
    for (; i <= n; i += i & -i) bit[i] += a;
  }
  int lower_bound(T k) {
    if (k <= 0) return 0;
    int ret = 0, i = 1;
    while ((i << 1) <= n) i <<= 1;
    for (; i; i >>= 1)
      if (ret + i <= n && bit[ret + i] < k) k -= bit[ret += i];
    return ret + 1;
  }
};
int N;
int main() {
  scanf("%d", &N);
  vector<int> X(N);
  vector<pair<int, int> > vs(N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &X[i]);
  }
  for (int i = 0; i < N; i++) {
    scanf("%d", &vs[i].first);
    vs[i].second = X[i];
  }
  sort(X.begin(), X.end());
  sort(vs.begin(), vs.end());
  reverse(vs.begin(), vs.end());
  BIT<long long> P(N);
  BIT<int> cnt(N);
  long long ans = 0;
  for (pair<int, int> p : vs) {
    int v = p.first, x = p.second;
    int id = lower_bound(X.begin(), X.end(), x) - X.begin() + 1;
    ans += P.sum(N) - P.sum(id) - (long long)x * (cnt.sum(N) - cnt.sum(id));
    cnt.add(id, 1);
    P.add(id, x);
  }
  printf("%lld\n", ans);
}

### Prompt

Your challenge is to write 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>
using namespace std;
template <typename T>
struct BIT {
  int n;
  vector<T> bit;
  BIT(int n_ = 0) : n(n_), bit(n_ + 1) {}
  T sum(int i) {
    T ans = 0;
    for (; i > 0; i -= i & -i) ans += bit[i];
    return ans;
  }
  void add(int i, T a) {
    if (i == 0) return;
    for (; i <= n; i += i & -i) bit[i] += a;
  }
  int lower_bound(T k) {
    if (k <= 0) return 0;
    int ret = 0, i = 1;
    while ((i << 1) <= n) i <<= 1;
    for (; i; i >>= 1)
      if (ret + i <= n && bit[ret + i] < k) k -= bit[ret += i];
    return ret + 1;
  }
};
int N;
int main() {
  scanf("%d", &N);
  vector<int> X(N);
  vector<pair<int, int> > vs(N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &X[i]);
  }
  for (int i = 0; i < N; i++) {
    scanf("%d", &vs[i].first);
    vs[i].second = X[i];
  }
  sort(X.begin(), X.end());
  sort(vs.begin(), vs.end());
  reverse(vs.begin(), vs.end());
  BIT<long long> P(N);
  BIT<int> cnt(N);
  long long ans = 0;
  for (pair<int, int> p : vs) {
    int v = p.first, x = p.second;
    int id = lower_bound(X.begin(), X.end(), x) - X.begin() + 1;
    ans += P.sum(N) - P.sum(id) - (long long)x * (cnt.sum(N) - cnt.sum(id));
    cnt.add(id, 1);
    P.add(id, x);
  }
  printf("%lld\n", ans);
}

```