#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long int mod = 1e9 + 7;
const long long int pinf = ((long long int)2e18);
const long long int ninf = ((long long int)-2e18);
bool isPrime(long long int n) {
  if (n <= 1) return false;
  for (long long int i = 2; i < n; i++)
    if (n % i == 0) return false;
  return true;
}
long long int fact(long long int n, long long int p) {
  if (n >= p) return 0;
  long long int result = 1;
  for (long long int i = 1; i <= n; i++) result = (result * i) % p;
  return result;
}
long long int maxSubArraySum(vector<long long int> a, long long int size) {
  long long int max_so_far = INT_MIN, max_ending_here = 0;
  for (long long int i = 0; i < size; i++) {
    max_ending_here = max_ending_here + a[i];
    if (max_so_far < max_ending_here) max_so_far = max_ending_here;
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max((long long int)0, max_so_far);
}
long long int findSubarraySum(vector<long long int> arr, long long int n,
                              long long int sum) {
  map<long long int, long long int> prevSum;
  long long int res = 0;
  long long int currsum = 0;
  long long int flag = 0;
  for (long long int i = 0; i < n; i++) {
    currsum += arr[i];
    if (currsum == sum) res++;
    prevSum[currsum]++;
    if (prevSum.find(currsum - sum) != prevSum.end())
      res += (prevSum[currsum - sum]);
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  {
    long long int d, m, x, pos, sum = 0, n, k, y, i, j, p, flag = 1;
    cin >> n;
    vector<pair<long long int, long long int> > a(n + 1);
    for (long long int i = 0; i < n; i++) {
      cin >> a[i].first;
    }
    for (long long int i = 0; i < n; i++) cin >> a[i].second;
    a[n] = {pinf, 0};
    sort(a.begin(), a.end());
    long long int ans = 0;
    priority_queue<long long int> q;
    q.push(a[0].second);
    long long int sz = a[0].second;
    for (long long int i = 1; i < n + 1; i++) {
      long long int c = a[i].first - a[i - 1].first;
      while ((!q.empty()) && (c)) {
        sz -= q.top();
        q.pop();
        ans += sz;
        c--;
      }
      q.push(a[i].second);
      sz += a[i].second;
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long int mod = 1e9 + 7;
const long long int pinf = ((long long int)2e18);
const long long int ninf = ((long long int)-2e18);
bool isPrime(long long int n) {
  if (n <= 1) return false;
  for (long long int i = 2; i < n; i++)
    if (n % i == 0) return false;
  return true;
}
long long int fact(long long int n, long long int p) {
  if (n >= p) return 0;
  long long int result = 1;
  for (long long int i = 1; i <= n; i++) result = (result * i) % p;
  return result;
}
long long int maxSubArraySum(vector<long long int> a, long long int size) {
  long long int max_so_far = INT_MIN, max_ending_here = 0;
  for (long long int i = 0; i < size; i++) {
    max_ending_here = max_ending_here + a[i];
    if (max_so_far < max_ending_here) max_so_far = max_ending_here;
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max((long long int)0, max_so_far);
}
long long int findSubarraySum(vector<long long int> arr, long long int n,
                              long long int sum) {
  map<long long int, long long int> prevSum;
  long long int res = 0;
  long long int currsum = 0;
  long long int flag = 0;
  for (long long int i = 0; i < n; i++) {
    currsum += arr[i];
    if (currsum == sum) res++;
    prevSum[currsum]++;
    if (prevSum.find(currsum - sum) != prevSum.end())
      res += (prevSum[currsum - sum]);
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  {
    long long int d, m, x, pos, sum = 0, n, k, y, i, j, p, flag = 1;
    cin >> n;
    vector<pair<long long int, long long int> > a(n + 1);
    for (long long int i = 0; i < n; i++) {
      cin >> a[i].first;
    }
    for (long long int i = 0; i < n; i++) cin >> a[i].second;
    a[n] = {pinf, 0};
    sort(a.begin(), a.end());
    long long int ans = 0;
    priority_queue<long long int> q;
    q.push(a[0].second);
    long long int sz = a[0].second;
    for (long long int i = 1; i < n + 1; i++) {
      long long int c = a[i].first - a[i - 1].first;
      while ((!q.empty()) && (c)) {
        sz -= q.top();
        q.pop();
        ans += sz;
        c--;
      }
      q.push(a[i].second);
      sz += a[i].second;
    }
    cout << ans << "\n";
  }
  return 0;
}

```