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 f(long long l, long long k) {
  long long ans = 0;
  long long md = l % k;
  ans = md * ((l / k + 1) * (l / k + 1));
  ans += (l / k) * (l / k) * (k - md);
  return ans;
}
int main() {
  iostream::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  multiset<array<long long, 3> > se;
  se.clear();
  for (int i = 0; i < n; i++) se.insert({f(a[i], 1) - f(a[i], 2), a[i], 2});
  while (k > n) {
    k--;
    array<long long, 3> cur = *se.rbegin();
    se.erase(se.find(cur));
    se.insert({f(cur[1], cur[2]) - f(cur[1], cur[2] + 1), cur[1], cur[2] + 1});
  }
  long long ans = 0;
  for (auto it : se) ans += f(it[1], it[2] - 1);
  cout << ans << '\n';
}
