#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
inline void print(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) print(x / 10);
  putchar(x % 10 + 48);
}
long long n;
struct node {
  long long a, t;
} a[200010];
bool cmp(node x, node y) { return x.t > y.t; }
set<long long> st;
map<long long, bool> mp;
set<long long>::iterator it;
long long ans;
signed main() {
  n = read();
  for (long long i = 1; i <= n; ++i) a[i].a = read();
  for (long long i = 1; i <= n; ++i) a[i].t = read();
  sort(a + 1, a + n + 1, cmp);
  for (long long i = 1; i <= n; ++i) {
    if (!mp.count(a[i].a)) {
      it = st.lower_bound(a[i].a);
      if ((*it) == a[i].a) st.erase(it);
      mp[a[i].a] = 1;
      if (!mp.count(a[i].a + 1)) st.insert(a[i].a + 1);
    } else {
      it = st.lower_bound(a[i].a);
      if (!mp.count((*it) + 1)) st.insert((*it) + 1);
      mp[(*it)] = 1;
      ans += ((*it) - a[i].a) * a[i].t;
      st.erase(it);
    }
  }
  printf("%lld\n", ans);
}

### 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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
inline void print(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) print(x / 10);
  putchar(x % 10 + 48);
}
long long n;
struct node {
  long long a, t;
} a[200010];
bool cmp(node x, node y) { return x.t > y.t; }
set<long long> st;
map<long long, bool> mp;
set<long long>::iterator it;
long long ans;
signed main() {
  n = read();
  for (long long i = 1; i <= n; ++i) a[i].a = read();
  for (long long i = 1; i <= n; ++i) a[i].t = read();
  sort(a + 1, a + n + 1, cmp);
  for (long long i = 1; i <= n; ++i) {
    if (!mp.count(a[i].a)) {
      it = st.lower_bound(a[i].a);
      if ((*it) == a[i].a) st.erase(it);
      mp[a[i].a] = 1;
      if (!mp.count(a[i].a + 1)) st.insert(a[i].a + 1);
    } else {
      it = st.lower_bound(a[i].a);
      if (!mp.count((*it) + 1)) st.insert((*it) + 1);
      mp[(*it)] = 1;
      ans += ((*it) - a[i].a) * a[i].t;
      st.erase(it);
    }
  }
  printf("%lld\n", ans);
}

```