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;
priority_queue<pair<pair<long long, int>, int> > q;
int n, k;
long long a[200005];
long long che(long long now, long long time1) {
  long long temp = now / time1;
  long long temp1 = now % (time1);
  return temp * temp * (time1 - temp1) + (temp + 1) * (temp + 1) * (temp1);
}
int main() {
  cin >> n >> k;
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i] * a[i];
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) continue;
    q.push(make_pair(make_pair(che(a[i], 1) - che(a[i], 2), a[i]), 2));
  }
  k -= n;
  while (k > 0) {
    long long temp1 = q.top().first.first;
    long long temp2 = q.top().first.second;
    long long temp3 = q.top().second;
    q.pop();
    sum -= temp1;
    k--;
    if (temp2 == temp3) continue;
    q.push(
        make_pair(make_pair(che(temp2, temp3) - che(temp2, temp3 + 1), temp2),
                  temp3 + 1));
  }
  cout << sum;
}
