#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 15;
int tot = 1, N;
const int L = -2e8;
const int R = 2e8;
struct Node {
  int l, r, num;
  long long sum;
  Node() { l = 0, r = 0, num = 0, sum = 0; }
} E[maxn * 40];
struct Point {
  long long x, v;
  bool operator<(const Point &T) const { return x > T.x; }
} P[maxn];
void update(int l, int r, int x, int v, int id) {
  if (l == r) {
    E[id].num++;
    E[id].sum += x;
    return;
  }
  int mid = (l + r) >> 1;
  if (v <= mid) {
    if (!E[id].l) E[id].l = ++tot;
    update(l, mid, x, v, E[id].l);
  } else {
    if (!E[id].r) E[id].r = ++tot;
    update(mid + 1, r, x, v, E[id].r);
  }
  E[id].num = E[E[id].l].num + E[E[id].r].num;
  E[id].sum = E[E[id].l].sum + E[E[id].r].sum;
}
void getSum(int l, int r, int ln, int rn, int id, int &num, long long &sum) {
  if (ln > rn) return;
  if (ln == l && rn == r) {
    num += E[id].num;
    sum += E[id].sum;
    return;
  }
  int mid = (l + r) >> 1;
  if (ln > mid && E[id].r)
    getSum(mid + 1, r, ln, rn, E[id].r, num, sum);
  else if (rn <= mid && E[id].l)
    getSum(l, mid, ln, rn, E[id].l, num, sum);
  else {
    if (E[id].r) getSum(mid + 1, r, mid + 1, rn, E[id].r, num, sum);
    if (E[id].l) getSum(l, mid, ln, mid, E[id].l, num, sum);
  }
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%lld", &P[i].x);
  for (int i = 1; i <= N; i++) scanf("%lld", &P[i].v);
  sort(P + 1, P + 1 + N);
  long long ans = 0;
  for (int i = 1; i <= N; i++) {
    int num = 0;
    long long sum = 0;
    getSum(L, R, P[i].v, R, 1, num, sum);
    ans += (sum - 1LL * num * P[i].x);
    update(L, R, P[i].x, P[i].v, 1);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Generate 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;
const int maxn = 2e5 + 15;
int tot = 1, N;
const int L = -2e8;
const int R = 2e8;
struct Node {
  int l, r, num;
  long long sum;
  Node() { l = 0, r = 0, num = 0, sum = 0; }
} E[maxn * 40];
struct Point {
  long long x, v;
  bool operator<(const Point &T) const { return x > T.x; }
} P[maxn];
void update(int l, int r, int x, int v, int id) {
  if (l == r) {
    E[id].num++;
    E[id].sum += x;
    return;
  }
  int mid = (l + r) >> 1;
  if (v <= mid) {
    if (!E[id].l) E[id].l = ++tot;
    update(l, mid, x, v, E[id].l);
  } else {
    if (!E[id].r) E[id].r = ++tot;
    update(mid + 1, r, x, v, E[id].r);
  }
  E[id].num = E[E[id].l].num + E[E[id].r].num;
  E[id].sum = E[E[id].l].sum + E[E[id].r].sum;
}
void getSum(int l, int r, int ln, int rn, int id, int &num, long long &sum) {
  if (ln > rn) return;
  if (ln == l && rn == r) {
    num += E[id].num;
    sum += E[id].sum;
    return;
  }
  int mid = (l + r) >> 1;
  if (ln > mid && E[id].r)
    getSum(mid + 1, r, ln, rn, E[id].r, num, sum);
  else if (rn <= mid && E[id].l)
    getSum(l, mid, ln, rn, E[id].l, num, sum);
  else {
    if (E[id].r) getSum(mid + 1, r, mid + 1, rn, E[id].r, num, sum);
    if (E[id].l) getSum(l, mid, ln, mid, E[id].l, num, sum);
  }
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%lld", &P[i].x);
  for (int i = 1; i <= N; i++) scanf("%lld", &P[i].v);
  sort(P + 1, P + 1 + N);
  long long ans = 0;
  for (int i = 1; i <= N; i++) {
    int num = 0;
    long long sum = 0;
    getSum(L, R, P[i].v, R, 1, num, sum);
    ans += (sum - 1LL * num * P[i].x);
    update(L, R, P[i].x, P[i].v, 1);
  }
  printf("%lld\n", ans);
  return 0;
}

```