#include <bits/stdc++.h>
#pragma optimize("O3")
using namespace std;
const long long MOD = 1e9 + 9;
const long long INF = 1e9 + 7;
const int base = 2e5 + 1;
const long long MAX = 1e15 + 1;
const double EPS = 1e-9;
const double PI = acos(-1.);
const int MAXN = 2 * 1e5 + 47;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
pair<long long, long long> t[MAXN];
map<int, int> ind, vx;
void add(int x, long long val) {
  for (int j = x; j < MAXN; j = j | (j + 1)) {
    t[j].first += val;
    t[j].second++;
  }
}
long long get(int x) {
  long long res = 0;
  for (int j = x; j >= 0; j = (j & (j + 1)) - 1) {
    res += t[j].first;
  }
  return res;
}
long long cnt(int x) {
  long long res = 0;
  for (int j = x; j >= 0; j = (j & (j + 1)) - 1) {
    res += t[j].second;
  }
  return res;
}
int main() {
  int n;
  cin >> n;
  vector<long long> x(n), v(n);
  for (long long i = (0); i < (n); ++i) scanf("%lld", &x[i]);
  for (long long i = (0); i < (n); ++i) {
    scanf("%lld", &v[i]);
    vx[x[i]] = v[i];
  }
  sort(v.begin(), v.end());
  sort(x.begin(), x.end());
  long long res = 0;
  for (long long i = (0); i < (n); ++i) ind[v[i]] = i;
  for (long long i = (0); i < (n); ++i) {
    int cur = ind[vx[x[i]]];
    res += x[i] * cnt(cur) - get(cur);
    add(cur, x[i]);
  }
  cout << res;
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
#pragma optimize("O3")
using namespace std;
const long long MOD = 1e9 + 9;
const long long INF = 1e9 + 7;
const int base = 2e5 + 1;
const long long MAX = 1e15 + 1;
const double EPS = 1e-9;
const double PI = acos(-1.);
const int MAXN = 2 * 1e5 + 47;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
pair<long long, long long> t[MAXN];
map<int, int> ind, vx;
void add(int x, long long val) {
  for (int j = x; j < MAXN; j = j | (j + 1)) {
    t[j].first += val;
    t[j].second++;
  }
}
long long get(int x) {
  long long res = 0;
  for (int j = x; j >= 0; j = (j & (j + 1)) - 1) {
    res += t[j].first;
  }
  return res;
}
long long cnt(int x) {
  long long res = 0;
  for (int j = x; j >= 0; j = (j & (j + 1)) - 1) {
    res += t[j].second;
  }
  return res;
}
int main() {
  int n;
  cin >> n;
  vector<long long> x(n), v(n);
  for (long long i = (0); i < (n); ++i) scanf("%lld", &x[i]);
  for (long long i = (0); i < (n); ++i) {
    scanf("%lld", &v[i]);
    vx[x[i]] = v[i];
  }
  sort(v.begin(), v.end());
  sort(x.begin(), x.end());
  long long res = 0;
  for (long long i = (0); i < (n); ++i) ind[v[i]] = i;
  for (long long i = (0); i < (n); ++i) {
    int cur = ind[vx[x[i]]];
    res += x[i] * cnt(cur) - get(cur);
    add(cur, x[i]);
  }
  cout << res;
  return 0;
}

```