#include <bits/stdc++.h>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,abm,mmx,avx,avx2,popcnt,fma")
#pragma GCC optimize("fast-math")
using namespace std;
const long long mod = 1e9 + 7, mod2 = 998244353;
const long double eps = 1e-6;
const long long inf = 1e18;
const int base = 1e9;
const long double PI = 3.1415926;
void deb(long double first){};
int n;
long long ans = 0;
long long a[200 * 1000], t[200 * 1000];
vector<pair<long long, long long>> v;
priority_queue<pair<long long, long long>> q;
unsigned int bitcount(unsigned long long val) {
  return __builtin_popcount(val) + __builtin_popcount(val >> 32);
}
void local() { freopen("test.in", "r", stdin); }
void remote() {}
void read() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> t[i];
}
int main() {
  remote();
  read();
  v.resize(n);
  for (int i = 0; i < n; i++) {
    v[i] = {a[i], t[i]};
  }
  sort(v.begin(), v.end());
  long long last = v[0].first;
  for (int i = 0; i < n; i++) {
    long long free = v[i].first - last, used = 0;
    while (!q.empty() && free > 0) {
      const pair<long long, long long> e = q.top();
      q.pop();
      ans += e.first * (used + last - e.second);
      free--, used++;
    }
    last = v[i].first;
    while (i < v.size() && v[i].first == last) {
      q.push({v[i].second, v[i].first});
      i++;
    }
    i--;
  }
  long long used = 0;
  while (!q.empty()) {
    const pair<long long, long long> e = q.top();
    q.pop();
    ans += e.first * (used + last - e.second);
    used++;
  }
  cout << ans << '\n';
  ;
}

### Prompt

Create a solution in Cpp for 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 target("sse,sse2,sse3,ssse3,sse4,abm,mmx,avx,avx2,popcnt,fma")
#pragma GCC optimize("fast-math")
using namespace std;
const long long mod = 1e9 + 7, mod2 = 998244353;
const long double eps = 1e-6;
const long long inf = 1e18;
const int base = 1e9;
const long double PI = 3.1415926;
void deb(long double first){};
int n;
long long ans = 0;
long long a[200 * 1000], t[200 * 1000];
vector<pair<long long, long long>> v;
priority_queue<pair<long long, long long>> q;
unsigned int bitcount(unsigned long long val) {
  return __builtin_popcount(val) + __builtin_popcount(val >> 32);
}
void local() { freopen("test.in", "r", stdin); }
void remote() {}
void read() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> t[i];
}
int main() {
  remote();
  read();
  v.resize(n);
  for (int i = 0; i < n; i++) {
    v[i] = {a[i], t[i]};
  }
  sort(v.begin(), v.end());
  long long last = v[0].first;
  for (int i = 0; i < n; i++) {
    long long free = v[i].first - last, used = 0;
    while (!q.empty() && free > 0) {
      const pair<long long, long long> e = q.top();
      q.pop();
      ans += e.first * (used + last - e.second);
      free--, used++;
    }
    last = v[i].first;
    while (i < v.size() && v[i].first == last) {
      q.push({v[i].second, v[i].first});
      i++;
    }
    i--;
  }
  long long used = 0;
  while (!q.empty()) {
    const pair<long long, long long> e = q.top();
    q.pop();
    ans += e.first * (used + last - e.second);
    used++;
  }
  cout << ans << '\n';
  ;
}

```