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;
int const Nmax = 1e5 + 50;
int n, k;
int64_t a[Nmax];
int curPieces[Nmax];
int64_t f(int id, int pieces) {
  int64_t w = a[id] / pieces;
  if (w == 0) return -1e18;
  int64_t k1 = a[id] % pieces;
  int64_t k2 = pieces - k1;
  return k1 * (w + 1) * (w + 1) + k2 * w * w;
}
void Enter() {
  cin >> n >> k;
  for (int i = (int)1; i <= (int)n; ++i) cin >> a[i];
  for (int i = (int)1; i <= (int)n; ++i) curPieces[i] = 1;
}
void Solve() {
  int64_t result = 0;
  for (int i = (int)1; i <= (int)n; ++i) result += (int64_t)a[i] * a[i];
  priority_queue<pair<int64_t, int> > pq;
  for (int i = (int)1; i <= (int)n; ++i)
    if (f(i, 2) != -1e18) pq.push(pair<int64_t, int>(f(i, 1) - f(i, 2), i));
  k -= n;
  while (k > 0) {
    pair<int64_t, int> TOP = pq.top();
    pq.pop();
    result -= TOP.first;
    curPieces[TOP.second]++;
    if (f(TOP.second, curPieces[TOP.second] + 1) != -1e18)
      pq.push(pair<int64_t, int>(f(TOP.second, curPieces[TOP.second]) -
                                     f(TOP.second, curPieces[TOP.second] + 1),
                                 TOP.second));
    k--;
  }
  cout << result;
}
int main() {
  {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  Enter();
  Solve();
}
