#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
template <typename T>
void ckmax(T &a, T b) {
  if (a < b) a = b;
}
template <typename T>
void ckmin(T &a, T b) {
  if (a > b) a = b;
}
inline int read() {
  int x = 0, neg = 1;
  char op = getchar();
  while (!isdigit(op)) {
    if (op == '-') neg = -1;
    op = getchar();
  }
  while (isdigit(op)) {
    x = 10 * x + op - '0';
    op = getchar();
  }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
const int N = 200005;
struct node {
  int x, y;
  friend bool operator<(const node &x, const node &y) { return x.x < y.x; }
} a[N];
int n;
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++) {
    scanf("%d", &a[i].x);
  }
  for (register int i = 1; i <= n; i++) {
    scanf("%d", &a[i].y);
  }
  sort(a + 1, a + n + 1);
  int cur = 0, cnt = 0;
  long long ans = 0ll, tmp = 0ll;
  priority_queue<int> pq;
  while (pq.size() || cur < n) {
    cnt++;
    if (pq.empty()) {
      cnt = a[cur + 1].x;
      while (cur < n && cnt == a[cur + 1].x) {
        tmp += a[++cur].y;
        pq.push(a[cur].y);
      }
    } else {
      while (cur < n && cnt == a[cur + 1].x) {
        tmp += a[++cur].y;
        pq.push(a[cur].y);
      }
    }
    tmp -= pq.top();
    pq.pop();
    ans += tmp;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.

The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds. 

What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.

Input

The first line of input consists of single integer n — the number of news categories (1 ≤ n ≤ 200 000).

The second line of input consists of n integers a_i — the number of publications of i-th category selected by the batch algorithm (1 ≤ a_i ≤ 10^9).

The third line of input consists of n integers t_i — time it takes for targeted algorithm to find one new publication of category i (1 ≤ t_i ≤ 10^5).

Output

Print one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.

Examples

Input


5
3 7 9 7 8
5 2 5 7 5


Output


6


Input


5
1 2 3 4 5
1 1 1 1 1


Output


0

Note

In the first example, it is possible to find three publications of the second type, which will take 6 seconds.

In the second example, all news categories contain a different number of publications.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
template <typename T>
void ckmax(T &a, T b) {
  if (a < b) a = b;
}
template <typename T>
void ckmin(T &a, T b) {
  if (a > b) a = b;
}
inline int read() {
  int x = 0, neg = 1;
  char op = getchar();
  while (!isdigit(op)) {
    if (op == '-') neg = -1;
    op = getchar();
  }
  while (isdigit(op)) {
    x = 10 * x + op - '0';
    op = getchar();
  }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
const int N = 200005;
struct node {
  int x, y;
  friend bool operator<(const node &x, const node &y) { return x.x < y.x; }
} a[N];
int n;
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++) {
    scanf("%d", &a[i].x);
  }
  for (register int i = 1; i <= n; i++) {
    scanf("%d", &a[i].y);
  }
  sort(a + 1, a + n + 1);
  int cur = 0, cnt = 0;
  long long ans = 0ll, tmp = 0ll;
  priority_queue<int> pq;
  while (pq.size() || cur < n) {
    cnt++;
    if (pq.empty()) {
      cnt = a[cur + 1].x;
      while (cur < n && cnt == a[cur + 1].x) {
        tmp += a[++cur].y;
        pq.push(a[cur].y);
      }
    } else {
      while (cur < n && cnt == a[cur + 1].x) {
        tmp += a[++cur].y;
        pq.push(a[cur].y);
      }
    }
    tmp -= pq.top();
    pq.pop();
    ans += tmp;
  }
  printf("%lld\n", ans);
  return 0;
}

```