#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T>
using V = vector<T>;
template <class T, class U>
using P = pair<T, U>;
using vll = V<ll>;
using vvll = V<vll>;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const ll MOD = 1000000007;
const ll HIGHINF = (ll)1e18;
const int INF = 1e9;
int n;
V<P<int, int> > at;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n;
  at.resize(n);
  for (int i = 0; i < n; i++) cin >> at[i].first;
  for (int i = 0; i < n; i++) cin >> at[i].second;
  sort(at.begin(), at.end());
  ll ans = 0;
  ll mc = 0;
  int preva = at[0].first;
  map<int, int> cost;
  for (int i = 0; i < n; i++) {
    if (i > 0) {
      for (int nxti = at[i - 1].first + 1; nxti < at[i].first && mc > 0;
           nxti++) {
        ans += mc;
        mc -= -(*cost.begin()).first;
        cost[(*cost.begin()).first]--;
        if ((*cost.begin()).second == 0) cost.erase((*cost.begin()).first);
      }
    }
    ans += mc;
    int j = i;
    for (; j < n && at[i].first == at[j].first; j++) {
      cost[-at[j].second]++;
      mc += at[j].second;
    }
    i = j - 1;
    mc -= -(*cost.begin()).first;
    cost[(*cost.begin()).first]--;
    if ((*cost.begin()).second == 0) cost.erase((*cost.begin()).first);
  }
  while (mc > 0) {
    ans += mc;
    mc -= -(*cost.begin()).first;
    cost[(*cost.begin()).first]--;
    if ((*cost.begin()).second == 0) cost.erase((*cost.begin()).first);
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
using ll = long long;
template <class T>
using V = vector<T>;
template <class T, class U>
using P = pair<T, U>;
using vll = V<ll>;
using vvll = V<vll>;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const ll MOD = 1000000007;
const ll HIGHINF = (ll)1e18;
const int INF = 1e9;
int n;
V<P<int, int> > at;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n;
  at.resize(n);
  for (int i = 0; i < n; i++) cin >> at[i].first;
  for (int i = 0; i < n; i++) cin >> at[i].second;
  sort(at.begin(), at.end());
  ll ans = 0;
  ll mc = 0;
  int preva = at[0].first;
  map<int, int> cost;
  for (int i = 0; i < n; i++) {
    if (i > 0) {
      for (int nxti = at[i - 1].first + 1; nxti < at[i].first && mc > 0;
           nxti++) {
        ans += mc;
        mc -= -(*cost.begin()).first;
        cost[(*cost.begin()).first]--;
        if ((*cost.begin()).second == 0) cost.erase((*cost.begin()).first);
      }
    }
    ans += mc;
    int j = i;
    for (; j < n && at[i].first == at[j].first; j++) {
      cost[-at[j].second]++;
      mc += at[j].second;
    }
    i = j - 1;
    mc -= -(*cost.begin()).first;
    cost[(*cost.begin()).first]--;
    if ((*cost.begin()).second == 0) cost.erase((*cost.begin()).first);
  }
  while (mc > 0) {
    ans += mc;
    mc -= -(*cost.begin()).first;
    cost[(*cost.begin()).first]--;
    if ((*cost.begin()).second == 0) cost.erase((*cost.begin()).first);
  }
  cout << ans << '\n';
  return 0;
}

```