#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
bool prime[1000001];
long long f[200005];
long long pow1(long long x, long long y) {
  long long res = 1;
  x = x % mod;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long divide(long long n) { return pow1(n, mod - 2); }
long long ncr(long long n, long long r) {
  if (n < r) return 0;
  return (f[n] * ((divide(f[r]) * divide(f[n - r])) % mod)) % mod;
}
void sieve() {
  memset(prime, true, sizeof(prime));
  for (long long i = 2; i * i <= 1000000; i++)
    if (prime[i])
      for (long long j = i * i; j <= 1000000; j += i) prime[j] = false;
  prime[0] = prime[1] = false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, i, j, x = 0, y = 0, ti = 0;
  cin >> n;
  vector<pair<long long, long long>> a(n);
  for (i = 0; i < n; i++) cin >> a[i].first;
  for (i = 0; i < n; i++) cin >> a[i].second;
  sort(a.begin(), a.end());
  multiset<long long, greater<long long>> m;
  j = 0;
  while (j < n || m.size()) {
    if (m.empty()) x = a[j].first;
    while (j < n && x == a[j].first) {
      y += a[j].second;
      m.insert(a[j].second);
      j++;
    }
    y -= *(m.begin());
    ti += y;
    m.erase(m.begin());
    x++;
  }
  cout << ti << endl;
  return 0;
}
void factorize(long long n, set<pair<long long, long long>> s) {
  long long count = 0;
  while (!(n % 2)) {
    n >>= 1;
    count++;
  }
  if (count) s.insert(make_pair(2, count));
  for (long long i = 3; i <= sqrt(n); i += 2) {
    count = 0;
    while (n % i == 0) {
      count++;
      n = n / i;
    }
    if (count) s.insert(make_pair(i, count));
  }
  if (n > 2) s.insert(make_pair(n, 1));
}

### Prompt

Develop a solution in cpp 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;
const long long mod = 998244353;
bool prime[1000001];
long long f[200005];
long long pow1(long long x, long long y) {
  long long res = 1;
  x = x % mod;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long divide(long long n) { return pow1(n, mod - 2); }
long long ncr(long long n, long long r) {
  if (n < r) return 0;
  return (f[n] * ((divide(f[r]) * divide(f[n - r])) % mod)) % mod;
}
void sieve() {
  memset(prime, true, sizeof(prime));
  for (long long i = 2; i * i <= 1000000; i++)
    if (prime[i])
      for (long long j = i * i; j <= 1000000; j += i) prime[j] = false;
  prime[0] = prime[1] = false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, i, j, x = 0, y = 0, ti = 0;
  cin >> n;
  vector<pair<long long, long long>> a(n);
  for (i = 0; i < n; i++) cin >> a[i].first;
  for (i = 0; i < n; i++) cin >> a[i].second;
  sort(a.begin(), a.end());
  multiset<long long, greater<long long>> m;
  j = 0;
  while (j < n || m.size()) {
    if (m.empty()) x = a[j].first;
    while (j < n && x == a[j].first) {
      y += a[j].second;
      m.insert(a[j].second);
      j++;
    }
    y -= *(m.begin());
    ti += y;
    m.erase(m.begin());
    x++;
  }
  cout << ti << endl;
  return 0;
}
void factorize(long long n, set<pair<long long, long long>> s) {
  long long count = 0;
  while (!(n % 2)) {
    n >>= 1;
    count++;
  }
  if (count) s.insert(make_pair(2, count));
  for (long long i = 3; i <= sqrt(n); i += 2) {
    count = 0;
    while (n % i == 0) {
      count++;
      n = n / i;
    }
    if (count) s.insert(make_pair(i, count));
  }
  if (n > 2) s.insert(make_pair(n, 1));
}

```