#include <bits/stdc++.h>
#pragma GСC optimize("Pivo na polu")
using namespace std;
const int mod = 1e9 + 7;
const int INF32 = 2e9;
const int SZ = 3e5 + 11;
const int logSZ = 20;
const long long INF64 = 4e18;
const long double eps = 1e-12;
const long long P = 53;
map<long long, long long> st;
set<long long> kek;
map<long long, vector<long long> > mp;
void Solve(int test_num) {
  long long n;
  cin >> n;
  vector<pair<long long, long long> > v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].first;
    kek.insert(v[i].first);
  }
  for (int i = 0; i < n; i++) {
    cin >> v[i].second;
    mp[v[i].first].push_back(v[i].second);
  }
  for (auto &i : mp) sort(i.second.begin(), i.second.end());
  long long i = ((*mp.begin()).first), ans = 0, lol = 0;
  while (true) {
    for (int j = mp[i].size() - 1; j >= 0; j--) {
      st[mp[i][j]]++;
      lol += mp[i][j];
    }
    if (!lol) {
      if (kek.upper_bound(i) == kek.end()) break;
      i = *kek.upper_bound(i);
      continue;
    }
    auto it = st.end();
    pair<long long, long long> jopa = *(--it);
    st[jopa.first]--;
    if (!st[jopa.first]) st.erase(jopa.first);
    lol -= jopa.first;
    ans += lol;
    i++;
  }
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  int q = 1;
  for (int i = 1; i <= q; i++) Solve(i);
}

### Prompt

In cpp, your task is to solve 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 GСC optimize("Pivo na polu")
using namespace std;
const int mod = 1e9 + 7;
const int INF32 = 2e9;
const int SZ = 3e5 + 11;
const int logSZ = 20;
const long long INF64 = 4e18;
const long double eps = 1e-12;
const long long P = 53;
map<long long, long long> st;
set<long long> kek;
map<long long, vector<long long> > mp;
void Solve(int test_num) {
  long long n;
  cin >> n;
  vector<pair<long long, long long> > v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].first;
    kek.insert(v[i].first);
  }
  for (int i = 0; i < n; i++) {
    cin >> v[i].second;
    mp[v[i].first].push_back(v[i].second);
  }
  for (auto &i : mp) sort(i.second.begin(), i.second.end());
  long long i = ((*mp.begin()).first), ans = 0, lol = 0;
  while (true) {
    for (int j = mp[i].size() - 1; j >= 0; j--) {
      st[mp[i][j]]++;
      lol += mp[i][j];
    }
    if (!lol) {
      if (kek.upper_bound(i) == kek.end()) break;
      i = *kek.upper_bound(i);
      continue;
    }
    auto it = st.end();
    pair<long long, long long> jopa = *(--it);
    st[jopa.first]--;
    if (!st[jopa.first]) st.erase(jopa.first);
    lol -= jopa.first;
    ans += lol;
    i++;
  }
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  int q = 1;
  for (int i = 1; i <= q; i++) Solve(i);
}

```