#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const long long N = 2 * 1e5 + 10;
const long long inf = 1e18;
const long long ninf = -1e18;
long long gcd(long long, long long);
long long modPower(long long, long long, long long);
long long power(long long, long long);
long long modInverse(long long, long long);
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long modPower(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = modPower(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long power(long long x, long long y) {
  if (y == 0) return 1;
  long long p = power(x, y / 2);
  p = (p * p);
  return (y % 2 == 0) ? p : (x * p);
}
long long modInverse(long long a, long long m) {
  return ((modPower(a, m - 2, m) + mod) % mod);
}
long long modDivide(long long a, long long b, long long m) {
  a = a % m;
  long long inv = modInverse(b, m);
  if (inv == -1)
    return 0;
  else
    return (inv * a) % m;
}
struct comparepq {
  bool operator()(pair<int, int> a, pair<int, int> b) {
    if (a.second == b.second) return a.first > b.first;
    return a.second < b.second;
  }
};
bool myCompare(pair<int, int> a, pair<int, int> b) {
  if (a.second == b.second) return a.first < b.first;
  return a.second > b.second;
}
void solve() {
  int n;
  cin >> n;
  pair<int, int> arr[n];
  for (long long i = (0); i < (n); i++) cin >> arr[i].first;
  for (long long i = (0); i < (n); i++) cin >> arr[i].second;
  sort(arr, arr + n);
  priority_queue<pair<int, int>, vector<pair<int, int> >, comparepq> pq;
  sort(arr, arr + n);
  int curr = arr[0].first, ptr = 0;
  long long ans = 0;
  while (ptr < n || pq.size()) {
    if (ptr < n) curr = arr[ptr].first;
    while (ptr < n && arr[ptr].first == curr) {
      pq.push(arr[ptr++]);
    }
    while ((ptr >= n || curr < arr[ptr].first) && !pq.empty()) {
      pair<int, int> p = pq.top();
      pq.pop();
      ans += (long long)p.second * (long long)(curr - p.first);
      curr++;
    }
  }
  cout << ans << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t = 1;
  for (long long i = (0); i < (t); i++) {
    solve();
  }
  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 mod = 998244353;
const long long N = 2 * 1e5 + 10;
const long long inf = 1e18;
const long long ninf = -1e18;
long long gcd(long long, long long);
long long modPower(long long, long long, long long);
long long power(long long, long long);
long long modInverse(long long, long long);
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long modPower(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = modPower(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long power(long long x, long long y) {
  if (y == 0) return 1;
  long long p = power(x, y / 2);
  p = (p * p);
  return (y % 2 == 0) ? p : (x * p);
}
long long modInverse(long long a, long long m) {
  return ((modPower(a, m - 2, m) + mod) % mod);
}
long long modDivide(long long a, long long b, long long m) {
  a = a % m;
  long long inv = modInverse(b, m);
  if (inv == -1)
    return 0;
  else
    return (inv * a) % m;
}
struct comparepq {
  bool operator()(pair<int, int> a, pair<int, int> b) {
    if (a.second == b.second) return a.first > b.first;
    return a.second < b.second;
  }
};
bool myCompare(pair<int, int> a, pair<int, int> b) {
  if (a.second == b.second) return a.first < b.first;
  return a.second > b.second;
}
void solve() {
  int n;
  cin >> n;
  pair<int, int> arr[n];
  for (long long i = (0); i < (n); i++) cin >> arr[i].first;
  for (long long i = (0); i < (n); i++) cin >> arr[i].second;
  sort(arr, arr + n);
  priority_queue<pair<int, int>, vector<pair<int, int> >, comparepq> pq;
  sort(arr, arr + n);
  int curr = arr[0].first, ptr = 0;
  long long ans = 0;
  while (ptr < n || pq.size()) {
    if (ptr < n) curr = arr[ptr].first;
    while (ptr < n && arr[ptr].first == curr) {
      pq.push(arr[ptr++]);
    }
    while ((ptr >= n || curr < arr[ptr].first) && !pq.empty()) {
      pair<int, int> p = pq.top();
      pq.pop();
      ans += (long long)p.second * (long long)(curr - p.first);
      curr++;
    }
  }
  cout << ans << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t = 1;
  for (long long i = (0); i < (t); i++) {
    solve();
  }
  return 0;
}

```