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 n, k;
long long a[100010];
struct node {
  long long id, x, y, step;
  bool operator<(const node& f) const { return x - y < f.x - f.y; }
};
long long ans = 0;
priority_queue<node> q;
void run(int T) {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    ans += a[i] * a[i];
    q.push((node){i, a[i] * a[i],
                  (a[i] / 2) * (a[i] / 2) * (2 - a[i] % 2) +
                      (a[i] / 2 + 1) * (a[i] / 2 + 1) * (a[i] % 2),
                  1});
  }
  for (int i = 1; i <= k - n; i++) {
    node top = q.top();
    q.pop();
    ans += top.y - top.x;
    long long j = top.step + 2;
    long long tmp = (a[top.id] / j) * (a[top.id] / j) * (j - a[top.id] % j) +
                    (a[top.id] / j + 1) * (a[top.id] / j + 1) * (a[top.id] % j);
    q.push((node){top.id, top.y, tmp, j - 1});
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(20);
  int T = 1;
  for (int i = 1; i <= T; i++) run(i);
}
