#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 50;
const long long maxn1 = 2e5 + 1;
const long long maxx = 1e13;
const long long mod = 1e9 + 7;
const long long mod1 = 998244353;
const long long mod2 = 1e8;
priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
               greater<pair<long long, long long>>>
    q;
vector<long long> adj[maxn];
pair<long long, long long> b[maxn];
pair<long long, pair<long long, long long>> c[maxn];
struct tk {
  string s;
  long long id;
};
bool lf1(tk a, tk b) { return (a.s < b.s || (a.s == b.s && a.id < b.id)); }
pair<long long, long long> a[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first;
  }
  for (int i = 1; i <= n; i++) cin >> a[i].second;
  sort(a + 1, a + n + 1);
  multiset<long long, greater<long long>> st;
  long long idx = 1;
  long long valnw = -1e15;
  long long sum = 0;
  long long ans = 0;
  while (1) {
    while (idx <= n && a[idx].first == valnw) {
      st.insert(a[idx].second);
      sum += a[idx].second;
      idx++;
    }
    if (!st.empty()) {
      sum -= *st.begin();
      st.erase(st.begin());
      valnw++;
      ans += sum;
    } else {
      if (idx <= n)
        valnw = a[idx].first;
      else
        break;
    }
  }
  cout << ans;
}

### 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>
using namespace std;
const long long maxn = 2e5 + 50;
const long long maxn1 = 2e5 + 1;
const long long maxx = 1e13;
const long long mod = 1e9 + 7;
const long long mod1 = 998244353;
const long long mod2 = 1e8;
priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
               greater<pair<long long, long long>>>
    q;
vector<long long> adj[maxn];
pair<long long, long long> b[maxn];
pair<long long, pair<long long, long long>> c[maxn];
struct tk {
  string s;
  long long id;
};
bool lf1(tk a, tk b) { return (a.s < b.s || (a.s == b.s && a.id < b.id)); }
pair<long long, long long> a[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first;
  }
  for (int i = 1; i <= n; i++) cin >> a[i].second;
  sort(a + 1, a + n + 1);
  multiset<long long, greater<long long>> st;
  long long idx = 1;
  long long valnw = -1e15;
  long long sum = 0;
  long long ans = 0;
  while (1) {
    while (idx <= n && a[idx].first == valnw) {
      st.insert(a[idx].second);
      sum += a[idx].second;
      idx++;
    }
    if (!st.empty()) {
      sum -= *st.begin();
      st.erase(st.begin());
      valnw++;
      ans += sum;
    } else {
      if (idx <= n)
        valnw = a[idx].first;
      else
        break;
    }
  }
  cout << ans;
}

```