#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5;
long long T, n, tot, ans;
struct number {
  long long a, b;
} num[N];
inline bool cmp(number a, number b) {
  return a.a < b.a || a.a == b.a && a.b < b.b;
}
priority_queue<long long> q;
void init(long long l, long long r) {
  for (register long long i = num[l].a, j = l, tot = 0; i <= num[l].a + r - l;
       ++i) {
    ans += tot;
    while (j <= r && i == num[j].a) q.push(num[j].b), tot += num[j].b, j++;
    tot -= q.top();
    q.pop();
  }
}
signed main() {
  scanf("%lld", &n);
  for (register long long i = 1; i <= n; ++i) scanf("%lld", &num[i].a);
  for (register long long i = 1; i <= n; ++i) scanf("%lld", &num[i].b);
  sort(num + 1, num + n + 1, cmp);
  long long i = 2;
  while (i <= n) {
    if (num[i].a != num[i - 1].a) {
      i++;
      continue;
    }
    long long sum = 1, now = num[i].a;
    long long j = i + 1;
    while (num[j].a <= now + sum && j <= n) sum++, j++;
    long long l, r;
    l = i - 1;
    r = j - 1;
    init(l, r);
    i = j;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for 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 long long N = 2e5 + 5;
long long T, n, tot, ans;
struct number {
  long long a, b;
} num[N];
inline bool cmp(number a, number b) {
  return a.a < b.a || a.a == b.a && a.b < b.b;
}
priority_queue<long long> q;
void init(long long l, long long r) {
  for (register long long i = num[l].a, j = l, tot = 0; i <= num[l].a + r - l;
       ++i) {
    ans += tot;
    while (j <= r && i == num[j].a) q.push(num[j].b), tot += num[j].b, j++;
    tot -= q.top();
    q.pop();
  }
}
signed main() {
  scanf("%lld", &n);
  for (register long long i = 1; i <= n; ++i) scanf("%lld", &num[i].a);
  for (register long long i = 1; i <= n; ++i) scanf("%lld", &num[i].b);
  sort(num + 1, num + n + 1, cmp);
  long long i = 2;
  while (i <= n) {
    if (num[i].a != num[i - 1].a) {
      i++;
      continue;
    }
    long long sum = 1, now = num[i].a;
    long long j = i + 1;
    while (num[j].a <= now + sum && j <= n) sum++, j++;
    long long l, r;
    l = i - 1;
    r = j - 1;
    init(l, r);
    i = j;
  }
  printf("%lld\n", ans);
  return 0;
}

```