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>
long long bestTime(long long total, long long num) {
  long long div = total / num;
  long long rem = total % num;
  long long nrem = num - total % num;
  long long res = nrem * div * div + rem * (div + 1) * (div + 1);
  return res;
}
int main() {
  long long n, k;
  scanf("%lld %lld", &n, &k);
  std::set<std::pair<long long, long long> > s;
  std::vector<long long> len(n, 0), ps(n, 1);
  long long ans(0);
  for (long p = 0; p < n; p++) {
    scanf("%lld", &len[p]);
    ans += len[p] * len[p];
    s.insert(std::make_pair(
        bestTime(len[p], ps[p] + 1) - bestTime(len[p], ps[p]), p));
  }
  for (long p = 0; p < k - n; p++) {
    std::pair<long long, long long> elem = *s.begin();
    long long improvement = elem.first;
    long long which = elem.second;
    ans += improvement;
    ++ps[which];
    s.erase(s.begin());
    s.insert(std::make_pair(
        bestTime(len[which], ps[which] + 1) - bestTime(len[which], ps[which]),
        which));
  }
  printf("%lld\n", ans);
  return 0;
}
