#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
const long long MAXN = 3e5 + 10, inf = 1e9 + 10;
long long n, m, ans;
struct node {
  long long a, t;
  friend bool operator<(node x, node y) { return x.a < y.a; }
} q[MAXN];
priority_queue<long long> f;
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) q[i].a = read();
  for (long long i = 1; i <= n; i++) q[i].t = read();
  for (long long i = 1; i <= n; i++) ans -= q[i].a * q[i].t;
  sort(q + 1, q + n + 1);
  long long p = 0, tot = 1;
  while (tot <= n || !f.empty()) {
    if (f.empty()) p = q[tot].a;
    while (tot <= n && q[tot].a <= p) {
      f.push(q[tot].t);
      tot++;
    }
    ans += p * f.top();
    p++;
    f.pop();
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
const long long MAXN = 3e5 + 10, inf = 1e9 + 10;
long long n, m, ans;
struct node {
  long long a, t;
  friend bool operator<(node x, node y) { return x.a < y.a; }
} q[MAXN];
priority_queue<long long> f;
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) q[i].a = read();
  for (long long i = 1; i <= n; i++) q[i].t = read();
  for (long long i = 1; i <= n; i++) ans -= q[i].a * q[i].t;
  sort(q + 1, q + n + 1);
  long long p = 0, tot = 1;
  while (tot <= n || !f.empty()) {
    if (f.empty()) p = q[tot].a;
    while (tot <= n && q[tot].a <= p) {
      f.push(q[tot].t);
      tot++;
    }
    ans += p * f.top();
    p++;
    f.pop();
  }
  printf("%lld\n", ans);
  return 0;
}

```