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 Rec {
  long long delta;
  int cur_segs;
  int id;
  friend bool operator<(const Rec &a, const Rec &b) {
    return a.delta < b.delta;
  }
  Rec(long long _delta, int _cur_segs, int _id)
      : delta(_delta), cur_segs(_cur_segs), id(_id) {}
};
int n, k;
long long a[100010];
priority_queue<Rec> Q;
long long calc(long long length, long long segs) {
  long long length_per_seg = length / segs;
  long long remaining = length % segs;
  return length_per_seg * length_per_seg * (segs - remaining) +
         (length_per_seg + 1) * (length_per_seg + 1) * remaining;
}
int main() {
  cin >> n >> k;
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    ans += calc(a[i], 1);
    if (a[i] >= 1 + 1) Q.emplace(calc(a[i], 1) - calc(a[i], 2), 1, i);
  }
  for (int i = n + 1; i <= k; ++i) {
    auto w = Q.top();
    Q.pop();
    ans -= w.delta;
    if (a[w.id] >= w.cur_segs + 2)
      Q.emplace(calc(a[w.id], w.cur_segs + 1) - calc(a[w.id], w.cur_segs + 2),
                w.cur_segs + 1, w.id);
  }
  cout << ans << endl;
  return 0;
}
