#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("no-stack-protector")
using namespace std;
const long long INF = 1e18;
const long long MAX_N = 200000 + 228;
long long si = 16;
pair<long long, long long> elem[MAX_N];
vector<long long> vec[MAX_N], vec2[MAX_N];
deque<long long> deq;
set<long long> num;
multiset<long long> now;
map<long long, vector<long long>> map1;
long long ans, sum;
long long a, b, c, d, e, f, p = 1000000007;
long long x, y;
signed main() {
  cin >> a;
  for (long long i = 0; i < a; i++) {
    cin >> c;
    elem[i].first = c;
    num.insert(c);
  }
  for (long long i = 0; i < a; i++) {
    cin >> d;
    elem[i].second = d;
    map1[elem[i].first].push_back(d);
  }
  while (num.size() > 0) {
    auto it = num.begin();
    for (auto i : map1[(*it)]) {
      now.insert(i);
      sum += i;
    }
    if (now.size() == 1) {
      now.clear();
      sum = 0;
    } else {
      auto it2 = now.end();
      it2--;
      sum -= (*it2);
      ans += sum;
      now.erase(it2);
      num.insert((*it) + 1);
    }
    num.erase((*it));
  }
  cout << ans;
  cin >> b;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>
#pragma GCC optimize("O3")
#pragma GCC optimize("no-stack-protector")
using namespace std;
const long long INF = 1e18;
const long long MAX_N = 200000 + 228;
long long si = 16;
pair<long long, long long> elem[MAX_N];
vector<long long> vec[MAX_N], vec2[MAX_N];
deque<long long> deq;
set<long long> num;
multiset<long long> now;
map<long long, vector<long long>> map1;
long long ans, sum;
long long a, b, c, d, e, f, p = 1000000007;
long long x, y;
signed main() {
  cin >> a;
  for (long long i = 0; i < a; i++) {
    cin >> c;
    elem[i].first = c;
    num.insert(c);
  }
  for (long long i = 0; i < a; i++) {
    cin >> d;
    elem[i].second = d;
    map1[elem[i].first].push_back(d);
  }
  while (num.size() > 0) {
    auto it = num.begin();
    for (auto i : map1[(*it)]) {
      now.insert(i);
      sum += i;
    }
    if (now.size() == 1) {
      now.clear();
      sum = 0;
    } else {
      auto it2 = now.end();
      it2--;
      sum -= (*it2);
      ans += sum;
      now.erase(it2);
      num.insert((*it) + 1);
    }
    num.erase((*it));
  }
  cout << ans;
  cin >> b;
  return 0;
}

```