#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e15 + 10;
const long double pi = 3.141592653589793238;
long long MOD = 998244353;
const long long N = 2e5 + 10;
long long n;
pair<long long, long long> a[N];
void del(multiset<long long>& s, long long x) {
  auto it = s.find(x);
  if (it != s.end()) s.erase(it);
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i].first;
  for (long long i = 0; i < n; i++) cin >> a[i].second;
  sort(a, a + n);
  long long ans = 0, cursum = 0;
  multiset<long long> curset;
  for (long long i = 0; i < n - 1; i++) {
    cursum += a[i].second;
    curset.insert(a[i].second);
    if (a[i].first == a[i + 1].first) continue;
    long long mx = 0;
    if (curset.size() > 0) mx = *curset.rbegin();
    ans += cursum - mx;
    cursum -= mx;
    del(curset, mx);
    long long x1 = a[i].first, x2 = a[i + 1].first;
    long long times = x2 - x1 - 1;
    while (times > 0 && curset.size() > 0) {
      ans += cursum - *curset.rbegin();
      cursum -= *curset.rbegin();
      del(curset, *curset.rbegin());
      times--;
    }
  }
  cursum += a[n - 1].second;
  curset.insert(a[n - 1].second);
  long long mx = 0;
  if (curset.size() > 0) mx = *curset.rbegin();
  ans += cursum - mx;
  cursum -= mx;
  del(curset, mx);
  long long times = inf;
  while (times > 0 && curset.size() > 0) {
    ans += cursum - *curset.rbegin();
    cursum -= *curset.rbegin();
    del(curset, *curset.rbegin());
    times--;
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
}

### 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>
using namespace std;
const long long inf = 1e15 + 10;
const long double pi = 3.141592653589793238;
long long MOD = 998244353;
const long long N = 2e5 + 10;
long long n;
pair<long long, long long> a[N];
void del(multiset<long long>& s, long long x) {
  auto it = s.find(x);
  if (it != s.end()) s.erase(it);
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i].first;
  for (long long i = 0; i < n; i++) cin >> a[i].second;
  sort(a, a + n);
  long long ans = 0, cursum = 0;
  multiset<long long> curset;
  for (long long i = 0; i < n - 1; i++) {
    cursum += a[i].second;
    curset.insert(a[i].second);
    if (a[i].first == a[i + 1].first) continue;
    long long mx = 0;
    if (curset.size() > 0) mx = *curset.rbegin();
    ans += cursum - mx;
    cursum -= mx;
    del(curset, mx);
    long long x1 = a[i].first, x2 = a[i + 1].first;
    long long times = x2 - x1 - 1;
    while (times > 0 && curset.size() > 0) {
      ans += cursum - *curset.rbegin();
      cursum -= *curset.rbegin();
      del(curset, *curset.rbegin());
      times--;
    }
  }
  cursum += a[n - 1].second;
  curset.insert(a[n - 1].second);
  long long mx = 0;
  if (curset.size() > 0) mx = *curset.rbegin();
  ans += cursum - mx;
  cursum -= mx;
  del(curset, mx);
  long long times = inf;
  while (times > 0 && curset.size() > 0) {
    ans += cursum - *curset.rbegin();
    cursum -= *curset.rbegin();
    del(curset, *curset.rbegin());
    times--;
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
}

```