#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const int M = 200000;
const int nmax = 200010;
const int MOD = (1e9) + 7;
int n;
pair<int, int> p[nmax];
void solve() {
  cin >> n;
  multiset<int> st;
  map<int, int> mp;
  map<int, vector<int>> vc;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].first;
    mp[p[i].first]++;
  }
  for (int i = 1; i <= n; i++) {
    cin >> p[i].second;
    vc[p[i].first].push_back(p[i].second);
  }
  long long int ans = 0;
  long long int cost = 0;
  while (!mp.empty()) {
    int x = mp.begin()->first;
    int y = mp.begin()->second;
    mp.erase(mp.begin());
    for (auto v : vc[x]) {
      st.insert(-v);
      cost += v;
    }
    vc[x].clear();
    if (!st.empty()) {
      cost += (*st.begin());
      ans = ans + cost;
      st.erase(st.begin());
      if (!st.empty() && y > 1) mp[x + 1] += (y - 1);
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  solve();
  return 0;
}

### Prompt

Your task is to create 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")
using namespace std;
const int M = 200000;
const int nmax = 200010;
const int MOD = (1e9) + 7;
int n;
pair<int, int> p[nmax];
void solve() {
  cin >> n;
  multiset<int> st;
  map<int, int> mp;
  map<int, vector<int>> vc;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].first;
    mp[p[i].first]++;
  }
  for (int i = 1; i <= n; i++) {
    cin >> p[i].second;
    vc[p[i].first].push_back(p[i].second);
  }
  long long int ans = 0;
  long long int cost = 0;
  while (!mp.empty()) {
    int x = mp.begin()->first;
    int y = mp.begin()->second;
    mp.erase(mp.begin());
    for (auto v : vc[x]) {
      st.insert(-v);
      cost += v;
    }
    vc[x].clear();
    if (!st.empty()) {
      cost += (*st.begin());
      ans = ans + cost;
      st.erase(st.begin());
      if (!st.empty() && y > 1) mp[x + 1] += (y - 1);
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  solve();
  return 0;
}

```