#include <bits/stdc++.h>
using namespace std;
const long long Z = (long long)3e3 + 228;
const long long N = (long long)2e5 + 228;
const long long INF = (long long)1e9 + 228;
const long long MOD = (long long)998244353;
const long long LLINF = (long long)1e18 + 228;
const long double eps = (long double)1e-6;
long long a[N], b[N], t[N];
map<long long, long long> cnt;
map<long long, multiset<long long> > mp;
set<long long> news;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    news.insert(a[i]);
    cnt[a[i]]++;
  }
  for (long long i = 1; i <= n; i++) {
    cin >> t[i];
    mp[a[i]].insert(t[i]);
  }
  news.insert(LLINF);
  long long ans = 0, cur = -INF;
  while (true) {
    long long it = *news.upper_bound(cur);
    if (it == LLINF) break;
    long long sum = 0;
    multiset<long long> q;
    for (auto& z : mp[it]) {
      if (cnt[it] == 1) break;
      cnt[it]--;
      q.insert(z);
      sum += z;
    }
    long long seen = 0;
    for (long long i = it + 1; q.size() != 0; i++) {
      seen++;
      ans += sum;
      if (mp.find(i) != mp.end()) {
        for (auto& z : mp[i]) {
          q.insert(z);
          sum += z;
        }
      }
      sum -= *--q.end();
      q.erase(--q.end());
    }
    cur = it + seen;
  }
  cout << ans;
  return 0;
}

### 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>
using namespace std;
const long long Z = (long long)3e3 + 228;
const long long N = (long long)2e5 + 228;
const long long INF = (long long)1e9 + 228;
const long long MOD = (long long)998244353;
const long long LLINF = (long long)1e18 + 228;
const long double eps = (long double)1e-6;
long long a[N], b[N], t[N];
map<long long, long long> cnt;
map<long long, multiset<long long> > mp;
set<long long> news;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    news.insert(a[i]);
    cnt[a[i]]++;
  }
  for (long long i = 1; i <= n; i++) {
    cin >> t[i];
    mp[a[i]].insert(t[i]);
  }
  news.insert(LLINF);
  long long ans = 0, cur = -INF;
  while (true) {
    long long it = *news.upper_bound(cur);
    if (it == LLINF) break;
    long long sum = 0;
    multiset<long long> q;
    for (auto& z : mp[it]) {
      if (cnt[it] == 1) break;
      cnt[it]--;
      q.insert(z);
      sum += z;
    }
    long long seen = 0;
    for (long long i = it + 1; q.size() != 0; i++) {
      seen++;
      ans += sum;
      if (mp.find(i) != mp.end()) {
        for (auto& z : mp[i]) {
          q.insert(z);
          sum += z;
        }
      }
      sum -= *--q.end();
      q.erase(--q.end());
    }
    cur = it + seen;
  }
  cout << ans;
  return 0;
}

```