#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int ch[N << 5][2] = {0}, tot = 1, trr[N << 5] = {0};
void update(int &id, int l, int r, int pos) {
  if (id == 0) id = ++tot;
  trr[id] += 1;
  if (l == r) return;
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(ch[id][0], l, mid, pos);
  else
    update(ch[id][1], mid + 1, r, pos);
}
int query(int id, int l, int r, int L, int R) {
  if (R < L) return 0;
  if (id == 0) return 0;
  if (L <= l && r <= R) return trr[id];
  int mid = (l + r) / 2;
  if (L >= mid + 1)
    return query(ch[id][1], mid + 1, r, L, R);
  else if (R <= mid)
    return query(ch[id][0], l, mid, L, R);
  else
    return query(ch[id][0], l, mid, L, mid) +
           query(ch[id][1], mid + 1, r, mid + 1, R);
}
struct node {
  int val, exp;
  friend bool operator<(struct node a, struct node b) { return a.exp > b.exp; }
} a[N];
int main() {
  int n, l, r, mid;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].val);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].exp);
  sort(a + 1, a + 1 + n);
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    l = a[i].val;
    r = 1e9 + 1e6;
    while (l <= r) {
      mid = (l + r) / 2;
      if (query(1, 1, 1e9 + 1e6, a[i].val, mid) != mid - a[i].val + 1) {
        if (query(1, 1, 1e9 + 1e6, a[i].val, mid - 1) == mid - a[i].val) break;
        r = mid - 1;
      } else
        l = mid + 1;
    }
    res += (long long)(mid - a[i].val) * a[i].exp;
    int rt = 1;
    update(rt, 1, 1e9 + 1e6, mid);
  }
  printf("%lld\n", res);
}

### Prompt

Your task is to create 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>
using namespace std;
const int N = 2e5 + 10;
int ch[N << 5][2] = {0}, tot = 1, trr[N << 5] = {0};
void update(int &id, int l, int r, int pos) {
  if (id == 0) id = ++tot;
  trr[id] += 1;
  if (l == r) return;
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(ch[id][0], l, mid, pos);
  else
    update(ch[id][1], mid + 1, r, pos);
}
int query(int id, int l, int r, int L, int R) {
  if (R < L) return 0;
  if (id == 0) return 0;
  if (L <= l && r <= R) return trr[id];
  int mid = (l + r) / 2;
  if (L >= mid + 1)
    return query(ch[id][1], mid + 1, r, L, R);
  else if (R <= mid)
    return query(ch[id][0], l, mid, L, R);
  else
    return query(ch[id][0], l, mid, L, mid) +
           query(ch[id][1], mid + 1, r, mid + 1, R);
}
struct node {
  int val, exp;
  friend bool operator<(struct node a, struct node b) { return a.exp > b.exp; }
} a[N];
int main() {
  int n, l, r, mid;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].val);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].exp);
  sort(a + 1, a + 1 + n);
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    l = a[i].val;
    r = 1e9 + 1e6;
    while (l <= r) {
      mid = (l + r) / 2;
      if (query(1, 1, 1e9 + 1e6, a[i].val, mid) != mid - a[i].val + 1) {
        if (query(1, 1, 1e9 + 1e6, a[i].val, mid - 1) == mid - a[i].val) break;
        r = mid - 1;
      } else
        l = mid + 1;
    }
    res += (long long)(mid - a[i].val) * a[i].exp;
    int rt = 1;
    update(rt, 1, 1e9 + 1e6, mid);
  }
  printf("%lld\n", res);
}

```