#include <bits/stdc++.h>
using namespace std;
namespace io {
const int SI = 1 << 21 | 1;
char IB[SI], *IS, *IT, OB[SI], *OS = OB, *OT = OS + SI - 1, c, ch[100];
int f, t;
inline void flush() { fwrite(OB, 1, OS - OB, stdout), OS = OB; }
inline void pc(char x) {
  *OS++ = x;
  if (OS == OT) flush();
}
template <class I>
inline void rd(I &x) {
  for (f = 1, c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                              IS == IT ? EOF : *IS++)
                            : *IS++);
       c < '0' || c > '9';
       c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++))
    if (c == '-') f = -1;
  for (x = 0; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + (c & 15),
      c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                      IS == IT ? EOF : *IS++)
                    : *IS++))
    ;
  x *= f;
}
inline void rds(char *s, int &x) {
  for (c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++);
       c < 33 || c > 126;
       c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++))
    ;
  for (x = 0; c >= 33 && c <= 126;
       s[++x] = c, c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                                   IS == IT ? EOF : *IS++)
                                 : *IS++))
    ;
  s[x + 1] = '\0';
}
template <class I>
inline void print(I x, char k = '\n') {
  if (!x) pc('0');
  if (x < 0) pc('-'), x = -x;
  while (x) ch[++t] = x % 10 + '0', x /= 10;
  while (t) pc(ch[t--]);
  pc(k);
}
inline void prints(string s) {
  int x = s.length();
  while (t < x) pc(s[t++]);
  pc('\n'), t = 0;
}
struct Flush {
  ~Flush() { flush(); }
} flusher;
}  // namespace io
using io::print;
using io::prints;
using io::rd;
using io::rds;
const int N = 2e5 + 7;
int n;
long long s, ans;
pair<long long, long long> p[N];
int main() {
  rd(n);
  for (int i = 1; i <= n; i++) rd(p[i].first);
  for (int i = 1; i <= n; i++) rd(p[i].second);
  sort(p + 1, p + n + 1);
  priority_queue<long long> q;
  q.push(p[1].second), s += p[1].second;
  for (int i = 2; i <= n; i++) {
    while (p[i - 1].first < p[i].first && q.size())
      ++p[i - 1].first, s -= q.top(), q.pop(), ans += s;
    q.push(p[i].second), s += p[i].second;
  }
  while (q.size()) s -= q.top(), q.pop(), ans += s;
  print(ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
namespace io {
const int SI = 1 << 21 | 1;
char IB[SI], *IS, *IT, OB[SI], *OS = OB, *OT = OS + SI - 1, c, ch[100];
int f, t;
inline void flush() { fwrite(OB, 1, OS - OB, stdout), OS = OB; }
inline void pc(char x) {
  *OS++ = x;
  if (OS == OT) flush();
}
template <class I>
inline void rd(I &x) {
  for (f = 1, c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                              IS == IT ? EOF : *IS++)
                            : *IS++);
       c < '0' || c > '9';
       c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++))
    if (c == '-') f = -1;
  for (x = 0; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + (c & 15),
      c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                      IS == IT ? EOF : *IS++)
                    : *IS++))
    ;
  x *= f;
}
inline void rds(char *s, int &x) {
  for (c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++);
       c < 33 || c > 126;
       c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++))
    ;
  for (x = 0; c >= 33 && c <= 126;
       s[++x] = c, c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                                   IS == IT ? EOF : *IS++)
                                 : *IS++))
    ;
  s[x + 1] = '\0';
}
template <class I>
inline void print(I x, char k = '\n') {
  if (!x) pc('0');
  if (x < 0) pc('-'), x = -x;
  while (x) ch[++t] = x % 10 + '0', x /= 10;
  while (t) pc(ch[t--]);
  pc(k);
}
inline void prints(string s) {
  int x = s.length();
  while (t < x) pc(s[t++]);
  pc('\n'), t = 0;
}
struct Flush {
  ~Flush() { flush(); }
} flusher;
}  // namespace io
using io::print;
using io::prints;
using io::rd;
using io::rds;
const int N = 2e5 + 7;
int n;
long long s, ans;
pair<long long, long long> p[N];
int main() {
  rd(n);
  for (int i = 1; i <= n; i++) rd(p[i].first);
  for (int i = 1; i <= n; i++) rd(p[i].second);
  sort(p + 1, p + n + 1);
  priority_queue<long long> q;
  q.push(p[1].second), s += p[1].second;
  for (int i = 2; i <= n; i++) {
    while (p[i - 1].first < p[i].first && q.size())
      ++p[i - 1].first, s -= q.top(), q.pop(), ans += s;
    q.push(p[i].second), s += p[i].second;
  }
  while (q.size()) s -= q.top(), q.pop(), ans += s;
  print(ans);
  return 0;
}

```