#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long sum = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (isdigit(ch)) sum = sum * 10 + (ch ^ 48), ch = getchar();
  return sum * ff;
}
const long long N = 2e5 + 5;
long long n, m, b[N], ans;
struct nood {
  long long ai, ti;
  inline bool friend operator<(const nood &c, const nood &d) {
    if (c.ai == d.ai) return c.ti < d.ti;
    return c.ai < d.ai;
  }
};
nood a[N];
inline void solve(long long l, long long r) {
  priority_queue<long long> q;
  long long tot = 0;
  for (long long i = a[l].ai, j = l; i <= a[l].ai + r - l; i++) {
    ans += tot;
    while (j <= r && i == a[j].ai) {
      q.push(a[j].ti);
      tot += a[j].ti;
      j++;
    }
    tot -= q.top();
    q.pop();
  }
}
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i].ai = read();
  for (long long i = 1; i <= n; i++) a[i].ti = read();
  sort(a + 1, a + n + 1);
  for (long long i = 1, j = 1; i <= n; i = j + 1) {
    j = i;
    while (j + 1 <= n && (j - i + 1) > (a[j + 1].ai - a[i].ai)) j++;
    solve(i, j);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
inline long long read() {
  long long sum = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (isdigit(ch)) sum = sum * 10 + (ch ^ 48), ch = getchar();
  return sum * ff;
}
const long long N = 2e5 + 5;
long long n, m, b[N], ans;
struct nood {
  long long ai, ti;
  inline bool friend operator<(const nood &c, const nood &d) {
    if (c.ai == d.ai) return c.ti < d.ti;
    return c.ai < d.ai;
  }
};
nood a[N];
inline void solve(long long l, long long r) {
  priority_queue<long long> q;
  long long tot = 0;
  for (long long i = a[l].ai, j = l; i <= a[l].ai + r - l; i++) {
    ans += tot;
    while (j <= r && i == a[j].ai) {
      q.push(a[j].ti);
      tot += a[j].ti;
      j++;
    }
    tot -= q.top();
    q.pop();
  }
}
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i].ai = read();
  for (long long i = 1; i <= n; i++) a[i].ti = read();
  sort(a + 1, a + n + 1);
  for (long long i = 1, j = 1; i <= n; i = j + 1) {
    j = i;
    while (j + 1 <= n && (j - i + 1) > (a[j + 1].ai - a[i].ai)) j++;
    solve(i, j);
  }
  printf("%lld\n", ans);
  return 0;
}

```