#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
const long long INF = 1e9 + 7;
const long long maxn = 2e5 + 7;
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  vector<long long> t(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < n; i++) {
    cin >> t[i];
  }
  vector<pair<long long, long long> > v(n);
  for (long long i = 0; i < n; i++) {
    v[i] = {a[i], t[i]};
  }
  sort(v.begin(), v.end());
  priority_queue<long long> pq;
  int prev = v[0].first;
  long long ans = 0;
  long long currSum = 0;
  for (long long i = 0; i < n; i++) {
    auto j = i;
    long long currNum = v[i].first;
    long long spacesBn = (currNum - (prev + 1));
    for (long long s = 0; s < spacesBn; s++) {
      if (pq.empty()) break;
      auto x = pq.top();
      pq.pop();
      currSum -= x;
      ans += (x * (s + 1));
    }
    ans += (currSum * (currNum - prev));
    while (j < n && v[j].first == v[i].first) {
      j++;
    }
    j--;
    if (i != j) {
      for (long long x = i; x < j; x++) {
        pq.push(v[x].second);
        currSum += v[x].second;
      }
    }
    if (!pq.empty() && pq.top() > v[j].second) {
      currSum -= pq.top();
      pq.pop();
      pq.push(v[j].second);
      currSum += v[j].second;
    }
    prev = currNum;
    i = j;
  }
  auto cost = 1;
  while (!pq.empty()) {
    auto x = pq.top();
    pq.pop();
    ans += (x * (cost));
    cost++;
  }
  cout << ans << '\n';
}
int main(int argc, char** argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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>
#pragma GCC optimize("Ofast")
const long long INF = 1e9 + 7;
const long long maxn = 2e5 + 7;
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  vector<long long> t(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < n; i++) {
    cin >> t[i];
  }
  vector<pair<long long, long long> > v(n);
  for (long long i = 0; i < n; i++) {
    v[i] = {a[i], t[i]};
  }
  sort(v.begin(), v.end());
  priority_queue<long long> pq;
  int prev = v[0].first;
  long long ans = 0;
  long long currSum = 0;
  for (long long i = 0; i < n; i++) {
    auto j = i;
    long long currNum = v[i].first;
    long long spacesBn = (currNum - (prev + 1));
    for (long long s = 0; s < spacesBn; s++) {
      if (pq.empty()) break;
      auto x = pq.top();
      pq.pop();
      currSum -= x;
      ans += (x * (s + 1));
    }
    ans += (currSum * (currNum - prev));
    while (j < n && v[j].first == v[i].first) {
      j++;
    }
    j--;
    if (i != j) {
      for (long long x = i; x < j; x++) {
        pq.push(v[x].second);
        currSum += v[x].second;
      }
    }
    if (!pq.empty() && pq.top() > v[j].second) {
      currSum -= pq.top();
      pq.pop();
      pq.push(v[j].second);
      currSum += v[j].second;
    }
    prev = currNum;
    i = j;
  }
  auto cost = 1;
  while (!pq.empty()) {
    auto x = pq.top();
    pq.pop();
    ans += (x * (cost));
    cost++;
  }
  cout << ans << '\n';
}
int main(int argc, char** argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```