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 int f(long long int l, long long int p) {
  long long int num1 = l / p;
  long long int num2 = l % p;
  return (p - num2) * (num1 * num1) + num2 * (num1 + 1) * (num1 + 1);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int n, k;
  cin >> n >> k;
  priority_queue<tuple<long long int, long long int, long long int>> q;
  long long int ans = 0;
  long long int dummy;
  for (long long int i = 0; i < n; i++) {
    cin >> dummy;
    ans += (dummy * dummy);
    q.push(make_tuple(f(dummy, 1) - f(dummy, 2), dummy, 2));
  }
  for (long long int i = 0; i < k - n; i++) {
    auto x = q.top();
    ans -= (get<0>(x));
    dummy = get<1>(x);
    long long int iter = get<2>(x);
    q.pop();
    q.push(make_tuple(f(dummy, iter) - f(dummy, iter + 1), dummy, iter + 1));
  }
  cout << ans << "\n";
}
