#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
long long int savev[maxn];
struct node {
  long long int x, v;
  node(){};
  node(long long int x, long long int v) : x(x), v(v){};
  bool operator<(const node& n) const { return x < n.x; }
} save[maxn];
map<long long int, long long int> getv;
long long int stree[maxn];
long long int vtree[maxn];
long long int n;
void add(long long int x, long long int d, long long int* tree) {
  while (x <= n) {
    tree[x] += d;
    x += (x & (-x));
  }
}
long long int sum(long long int x, long long int* tree) {
  long long int sum = 0;
  while (x > 0) {
    sum += tree[x];
    x -= (x & (-x));
  }
  return sum;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &save[i].x);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &save[i].v);
    savev[i] = save[i].v;
  }
  sort(savev + 1, savev + 1 + n);
  sort(save + 1, save + 1 + n);
  int vcnt = 0;
  for (int i = 1; i <= n; i++) {
    if (!getv[savev[i]]) {
      getv[savev[i]] = ++vcnt;
    }
  }
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    node& nownode = save[i];
    long long int nowv = getv[nownode.v];
    long long int sumdis = sum(nowv, vtree);
    long long int sumsmall = sum(nowv, stree);
    ans += nownode.x * sumsmall - sumdis;
    add(nowv, nownode.x, vtree);
    add(nowv, 1, stree);
  }
  printf("%lld", ans);
}

### 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 maxn = 2e5 + 10;
long long int savev[maxn];
struct node {
  long long int x, v;
  node(){};
  node(long long int x, long long int v) : x(x), v(v){};
  bool operator<(const node& n) const { return x < n.x; }
} save[maxn];
map<long long int, long long int> getv;
long long int stree[maxn];
long long int vtree[maxn];
long long int n;
void add(long long int x, long long int d, long long int* tree) {
  while (x <= n) {
    tree[x] += d;
    x += (x & (-x));
  }
}
long long int sum(long long int x, long long int* tree) {
  long long int sum = 0;
  while (x > 0) {
    sum += tree[x];
    x -= (x & (-x));
  }
  return sum;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &save[i].x);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &save[i].v);
    savev[i] = save[i].v;
  }
  sort(savev + 1, savev + 1 + n);
  sort(save + 1, save + 1 + n);
  int vcnt = 0;
  for (int i = 1; i <= n; i++) {
    if (!getv[savev[i]]) {
      getv[savev[i]] = ++vcnt;
    }
  }
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    node& nownode = save[i];
    long long int nowv = getv[nownode.v];
    long long int sumdis = sum(nowv, vtree);
    long long int sumsmall = sum(nowv, stree);
    ans += nownode.x * sumsmall - sumdis;
    add(nowv, nownode.x, vtree);
    add(nowv, 1, stree);
  }
  printf("%lld", ans);
}

```