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;
struct S {
  int64_t v;
  int64_t c;
};
int64_t val(const S& v) {
  int64_t d = v.v / v.c;
  int64_t r = v.v % v.c;
  int64_t ans = 0;
  ans += r * (d + 1) * (d + 1);
  ans += (v.c - r) * d * d;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(20);
  int k, n;
  cin >> k >> n;
  auto cmp = [](const S& lhs, const S& rhs) {
    int64_t lval = val(lhs) - val({lhs.v, lhs.c + 1});
    int64_t rval = val(rhs) - val({rhs.v, rhs.c + 1});
    return lval > rval;
  };
  multiset<S, decltype(cmp)> vals(cmp);
  for (decltype(k) repeat_index = 0; repeat_index < k; repeat_index++) {
    int64_t v;
    cin >> v;
    vals.insert({v, 1});
  }
  for (decltype(n - k) repeat_index = 0; repeat_index < n - k; repeat_index++) {
    auto it = vals.begin();
    auto v = *it;
    vals.erase(it);
    v.c++;
    vals.insert(v);
  }
  int64_t ans = 0;
  for (auto& v : vals) {
    ans += val(v);
  }
  cout << ans << endl;
  return 0;
}
