There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.

Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.

Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.

Input

The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.

The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.

It is guaranteed that the sum of a_i is at least k.

Output

Output one integer: the minimum sum of time taken for rabbits to eat carrots.

Examples

Input


3 6
5 3 1


Output


15


Input


1 4
19


Output


91

Note

For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15

For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91.
#include <bits/stdc++.h>
using namespace std;
long long calculate(long long n, long long x) {
  if (n < x) return 1e18;
  return (n % x) * (n / x + 1) * (n / x + 1) + (x - n % x) * (n / x) * (n / x);
}
int main() {
  long long int n, k;
  cin >> n >> k;
  long long int arr[1000000];
  vector<long long> parts(n);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    parts[i] = 1;
  }
  priority_queue<pair<long long, long long int>> q;
  for (int i = 0; i < n; i++) {
    q.push(make_pair(
        calculate(arr[i], parts[i]) - calculate(arr[i], parts[i] + 1), i));
  }
  for (int i = 0; i < k - n; i++) {
    auto x = q.top();
    q.pop();
    long long index = x.second;
    parts[index]++;
    q.push(make_pair(calculate(arr[index], parts[index]) -
                         calculate(arr[index], parts[index] + 1),
                     index));
  }
  for (int i = 0; i < n; i++) {
    ans += calculate(arr[i], parts[i]);
  }
  cout << ans << endl;
}
