#include <bits/stdc++.h>
using namespace std;
const long long MXN = 300 + 10;
const long long MX5 = 2e5 + 10;
const long long MX6 = 1e6 + 10;
const long long LOG = 20;
const long long INF = 1e15;
const double eps = 1e-9;
const long long MOD = 1e9 + 7;
long long power(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2, md) % md
                      : power(a * a % md, b / 2, md) % md));
}
long long bmm(long long a, long long b) {
  return (a % b == 0 ? b : bmm(b, a % b));
}
string base2(long long n) {
  string a = "";
  while (n >= 2) {
    a += (char)(n % 2 + '0');
    n /= 2;
  }
  a += (char)(n + '0');
  reverse((a).begin(), (a).end());
  return a;
}
long long n, ans;
long long A[MX5], T[MX5], Ans[MX5];
multiset<pair<long long, long long> > st[MX5], nw;
vector<long long> vec;
inline int GetId(int x) {
  return lower_bound(vec.begin(), vec.end(), x) - vec.begin() + 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> A[i], vec.push_back(A[i]);
  for (int i = 0; i < n; i++) cin >> T[i];
  vec.push_back(INF);
  sort(((vec)).begin(), ((vec)).end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  for (int i = 0; i < n; i++) {
    st[GetId(A[i])].insert({-T[i], i});
  }
  for (int i = 0; i < vec.size() - 1; i++) {
    long long now = vec[i], nxt = vec[i + 1];
    for (auto x : st[GetId(now)]) {
      nw.insert(x);
    }
    auto itr1 = nw.begin();
    Ans[itr1->second] = now;
    nw.erase(itr1);
    long long d = min(nxt - now - 1, (long long)nw.size()), tt = 1;
    while (tt <= d) {
      auto itr = nw.begin();
      Ans[itr->second] = tt + now;
      nw.erase(itr);
      tt++;
    }
  }
  for (int i = 0; i < n; i++) {
    ans += (Ans[i] - A[i]) * T[i];
  }
  cout << ans;
  return 0;
}

### 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;
const long long MXN = 300 + 10;
const long long MX5 = 2e5 + 10;
const long long MX6 = 1e6 + 10;
const long long LOG = 20;
const long long INF = 1e15;
const double eps = 1e-9;
const long long MOD = 1e9 + 7;
long long power(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2, md) % md
                      : power(a * a % md, b / 2, md) % md));
}
long long bmm(long long a, long long b) {
  return (a % b == 0 ? b : bmm(b, a % b));
}
string base2(long long n) {
  string a = "";
  while (n >= 2) {
    a += (char)(n % 2 + '0');
    n /= 2;
  }
  a += (char)(n + '0');
  reverse((a).begin(), (a).end());
  return a;
}
long long n, ans;
long long A[MX5], T[MX5], Ans[MX5];
multiset<pair<long long, long long> > st[MX5], nw;
vector<long long> vec;
inline int GetId(int x) {
  return lower_bound(vec.begin(), vec.end(), x) - vec.begin() + 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> A[i], vec.push_back(A[i]);
  for (int i = 0; i < n; i++) cin >> T[i];
  vec.push_back(INF);
  sort(((vec)).begin(), ((vec)).end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  for (int i = 0; i < n; i++) {
    st[GetId(A[i])].insert({-T[i], i});
  }
  for (int i = 0; i < vec.size() - 1; i++) {
    long long now = vec[i], nxt = vec[i + 1];
    for (auto x : st[GetId(now)]) {
      nw.insert(x);
    }
    auto itr1 = nw.begin();
    Ans[itr1->second] = now;
    nw.erase(itr1);
    long long d = min(nxt - now - 1, (long long)nw.size()), tt = 1;
    while (tt <= d) {
      auto itr = nw.begin();
      Ans[itr->second] = tt + now;
      nw.erase(itr);
      tt++;
    }
  }
  for (int i = 0; i < n; i++) {
    ans += (Ans[i] - A[i]) * T[i];
  }
  cout << ans;
  return 0;
}

```