#include <bits/stdc++.h>
using namespace std;
long long mod = (1000000007LL);
inline long long poww(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
inline pair<long long, long long> operator+(pair<long long, long long> A,
                                            pair<long long, long long> B) {
  return pair<long long, long long>(A.first + B.first, A.second + B.second);
}
inline pair<long long, long long> operator-(pair<long long, long long> A,
                                            pair<long long, long long> B) {
  return pair<long long, long long>(A.first - B.first, A.second - B.second);
}
inline long long mult(long long a, long long b) { return (a * b) % mod; }
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long n, m, q, k, v[300050];
long long ans;
long long a[300050];
pair<long long, long long> TT, w[300050];
string second;
long long id[300050];
map<long long, long long> maior, sum, qtd;
map<long long, vector<long long> > caras;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  set<long long> ss;
  for (long long i = 1; i <= n; i++) {
    cin >> v[i];
    ss.insert(a[i]);
    if (!maior[a[i]])
      maior[a[i]] = v[i];
    else
      maior[a[i]] = max(maior[a[i]], v[i]);
    sum[a[i]] += v[i];
    qtd[a[i]]++;
    caras[a[i]].push_back(v[i]);
  }
  priority_queue<long long> pq;
  long long curr = 0;
  for (auto dia : ss) {
    for (auto w : caras[dia]) pq.push(w);
    curr += sum[dia];
    if ((long long)(pq).size() > 1) {
      curr -= pq.top();
      pq.pop();
      ans += curr;
      ss.insert(dia + 1);
    } else
      curr -= pq.top(), pq.pop();
  }
  cout << ans << "\n";
}

### Prompt

Please create a solution in CPP 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>
using namespace std;
long long mod = (1000000007LL);
inline long long poww(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
inline pair<long long, long long> operator+(pair<long long, long long> A,
                                            pair<long long, long long> B) {
  return pair<long long, long long>(A.first + B.first, A.second + B.second);
}
inline pair<long long, long long> operator-(pair<long long, long long> A,
                                            pair<long long, long long> B) {
  return pair<long long, long long>(A.first - B.first, A.second - B.second);
}
inline long long mult(long long a, long long b) { return (a * b) % mod; }
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long n, m, q, k, v[300050];
long long ans;
long long a[300050];
pair<long long, long long> TT, w[300050];
string second;
long long id[300050];
map<long long, long long> maior, sum, qtd;
map<long long, vector<long long> > caras;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  set<long long> ss;
  for (long long i = 1; i <= n; i++) {
    cin >> v[i];
    ss.insert(a[i]);
    if (!maior[a[i]])
      maior[a[i]] = v[i];
    else
      maior[a[i]] = max(maior[a[i]], v[i]);
    sum[a[i]] += v[i];
    qtd[a[i]]++;
    caras[a[i]].push_back(v[i]);
  }
  priority_queue<long long> pq;
  long long curr = 0;
  for (auto dia : ss) {
    for (auto w : caras[dia]) pq.push(w);
    curr += sum[dia];
    if ((long long)(pq).size() > 1) {
      curr -= pq.top();
      pq.pop();
      ans += curr;
      ss.insert(dia + 1);
    } else
      curr -= pq.top(), pq.pop();
  }
  cout << ans << "\n";
}

```