#include <bits/stdc++.h>
using namespace std;
const int N = 200000 + 10;
struct Item {
  int a, t;
};
struct Node {
  int left, right, cnt;
};
Item item[N];
Node sgt[N * 20];
int tot;
bool cmp(const Item& a, const Item& b) { return a.t < b.t; }
int insert(int root, int begin, int end, int& x) {
  int mid, tmp;
  if (!root) root = ++tot;
  if (begin == end) {
    if (sgt[root].cnt == 0) sgt[root].cnt++;
    return root;
  }
  mid = (begin + end) / 2;
  if (x <= mid &&
      (!sgt[root].left || sgt[sgt[root].left].cnt < mid - begin + 1)) {
    tmp = sgt[sgt[root].left].cnt;
    sgt[root].left = insert(sgt[root].left, begin, mid, x);
    if (sgt[sgt[root].left].cnt == tmp) {
      x = max(x, mid + 1);
      sgt[root].right = insert(sgt[root].right, mid + 1, end, x);
    }
  } else {
    x = max(x, mid + 1);
    sgt[root].right = insert(sgt[root].right, mid + 1, end, x);
  }
  sgt[root].cnt = sgt[sgt[root].left].cnt + sgt[sgt[root].right].cnt;
  return root;
}
int main() {
  long long res;
  int n, x, y, i;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &item[i].a);
  for (i = 0; i < n; i++) scanf("%d", &item[i].t);
  sort(item, item + n, cmp);
  res = 0;
  tot = 1;
  for (i = n - 1; i >= 0; i--) {
    y = x = item[i].a;
    insert(1, 1, 1000000000 + N, y);
    res += 1LL * item[i].t * (y - x);
  }
  printf("%lld\n", res);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
using namespace std;
const int N = 200000 + 10;
struct Item {
  int a, t;
};
struct Node {
  int left, right, cnt;
};
Item item[N];
Node sgt[N * 20];
int tot;
bool cmp(const Item& a, const Item& b) { return a.t < b.t; }
int insert(int root, int begin, int end, int& x) {
  int mid, tmp;
  if (!root) root = ++tot;
  if (begin == end) {
    if (sgt[root].cnt == 0) sgt[root].cnt++;
    return root;
  }
  mid = (begin + end) / 2;
  if (x <= mid &&
      (!sgt[root].left || sgt[sgt[root].left].cnt < mid - begin + 1)) {
    tmp = sgt[sgt[root].left].cnt;
    sgt[root].left = insert(sgt[root].left, begin, mid, x);
    if (sgt[sgt[root].left].cnt == tmp) {
      x = max(x, mid + 1);
      sgt[root].right = insert(sgt[root].right, mid + 1, end, x);
    }
  } else {
    x = max(x, mid + 1);
    sgt[root].right = insert(sgt[root].right, mid + 1, end, x);
  }
  sgt[root].cnt = sgt[sgt[root].left].cnt + sgt[sgt[root].right].cnt;
  return root;
}
int main() {
  long long res;
  int n, x, y, i;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &item[i].a);
  for (i = 0; i < n; i++) scanf("%d", &item[i].t);
  sort(item, item + n, cmp);
  res = 0;
  tot = 1;
  for (i = n - 1; i >= 0; i--) {
    y = x = item[i].a;
    insert(1, 1, 1000000000 + N, y);
    res += 1LL * item[i].t * (y - x);
  }
  printf("%lld\n", res);
  return 0;
}

```