#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using pii = pair<int, int>;
static int MAX_N = 200000, n;
vector<ll> bitx(MAX_N + 1), bit1(MAX_N + 1);
ll sum(int i, vector<ll> &bit) {
  ll res = 0;
  while (i > 0) {
    res += bit.at(i);
    i -= i & -i;
  }
  return res;
}
void add(int i, int x, vector<ll> &bit) {
  while (i <= n) {
    bit.at(i) += x;
    i += i & -i;
  }
}
int main() {
  cin >> n;
  vi x(n);
  for (int i = 0; i < (int)(n); i++) cin >> x.at(i);
  vi v(n);
  for (int i = 0; i < (int)(n); i++) cin >> v.at(i);
  vector<pii> p;
  for (int i = 0; i < (int)(n); i++) {
    pii pi(x.at(i), v.at(i));
    p.push_back(pi);
  }
  sort(p.begin(), p.end());
  map<int, int> x2i;
  for (int i = 0; i < (int)(n); i++) x2i[p.at(i).first] = i + 1;
  sort(p.begin(), p.end(), [](pii p1, pii p2) {
    if (p1.second == p2.second)
      return p1.first < p2.first;
    else
      return p1.second < p2.second;
  });
  ll ans = 0;
  for (int j = 0; j < (int)(n); j++) {
    int x = p.at(j).first;
    int i = x2i.at(x);
    ans += x * sum(i - 1, bit1) - sum(i - 1, bitx);
    add(i, x, bitx);
    add(i, 1, bit1);
  }
  cout << ans << endl;
}

### Prompt

Your task is to create 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;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using pii = pair<int, int>;
static int MAX_N = 200000, n;
vector<ll> bitx(MAX_N + 1), bit1(MAX_N + 1);
ll sum(int i, vector<ll> &bit) {
  ll res = 0;
  while (i > 0) {
    res += bit.at(i);
    i -= i & -i;
  }
  return res;
}
void add(int i, int x, vector<ll> &bit) {
  while (i <= n) {
    bit.at(i) += x;
    i += i & -i;
  }
}
int main() {
  cin >> n;
  vi x(n);
  for (int i = 0; i < (int)(n); i++) cin >> x.at(i);
  vi v(n);
  for (int i = 0; i < (int)(n); i++) cin >> v.at(i);
  vector<pii> p;
  for (int i = 0; i < (int)(n); i++) {
    pii pi(x.at(i), v.at(i));
    p.push_back(pi);
  }
  sort(p.begin(), p.end());
  map<int, int> x2i;
  for (int i = 0; i < (int)(n); i++) x2i[p.at(i).first] = i + 1;
  sort(p.begin(), p.end(), [](pii p1, pii p2) {
    if (p1.second == p2.second)
      return p1.first < p2.first;
    else
      return p1.second < p2.second;
  });
  ll ans = 0;
  for (int j = 0; j < (int)(n); j++) {
    int x = p.at(j).first;
    int i = x2i.at(x);
    ans += x * sum(i - 1, bit1) - sum(i - 1, bitx);
    add(i, x, bitx);
    add(i, 1, bit1);
  }
  cout << ans << endl;
}

```