#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 6;
const long long M = 1e9 + 7;
map<long long, long long> mp;
vector<pair<long long, long long>> v;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.second == b.second)
    return a.first < b.first;
  else
    return a.second > b.second;
}
long long find(long long n) {
  mp[n] = mp[n] == 0 ? n : find(mp[n]);
  return mp[n];
}
void merge(long long a, long long b) {
  long long fa = find(a);
  mp[b] = fa;
}
int main() {
  long long n, x, y;
  scanf("%lld", &n);
  long long a[n];
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &y);
    v.push_back({a[i], y});
  }
  sort(v.begin(), v.end(), comp);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    x = v[i].first;
    y = v[i].second;
    long long rh = find(x);
    ans += (rh - x) * y;
    mp[rh] = rh + 1;
    mp[x] = rh + 1;
  }
  printf("%lld\n", ans);
  ;
}

### Prompt

Please create a solution in Cpp 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 + 6;
const long long M = 1e9 + 7;
map<long long, long long> mp;
vector<pair<long long, long long>> v;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.second == b.second)
    return a.first < b.first;
  else
    return a.second > b.second;
}
long long find(long long n) {
  mp[n] = mp[n] == 0 ? n : find(mp[n]);
  return mp[n];
}
void merge(long long a, long long b) {
  long long fa = find(a);
  mp[b] = fa;
}
int main() {
  long long n, x, y;
  scanf("%lld", &n);
  long long a[n];
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &y);
    v.push_back({a[i], y});
  }
  sort(v.begin(), v.end(), comp);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    x = v[i].first;
    y = v[i].second;
    long long rh = find(x);
    ans += (rh - x) * y;
    mp[rh] = rh + 1;
    mp[x] = rh + 1;
  }
  printf("%lld\n", ans);
  ;
}

```