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 main() {
  long long n, k;
  cin >> n >> k;
  vector<long long> arr(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  priority_queue<tuple<long long, int, long long> > pq;
  for (int i = 1; i <= n; i++) {
    long long num1 = arr[i] / 2;
    long long num2 = arr[i] % 2;
    long long var = arr[i] * arr[i] - num2 * (num1 + 1) * (num1 + 1) -
                    (2 - num2) * num1 * num1;
    pq.push(make_tuple(var, 1, i));
  }
  int cnt = n;
  while (cnt < k) {
    auto var = pq.top();
    pq.pop();
    long long a = get<0>(var);
    long long b = get<1>(var);
    long long c = get<2>(var);
    long long num1 = arr[c] / (b + 1);
    long long rem1 = arr[c] % (b + 1);
    long long num2 = arr[c] / (b + 2);
    long long rem2 = arr[c] % (b + 2);
    long long sum1 =
        rem1 * (num1 + 1) * (num1 + 1) + (b + 1 - rem1) * num1 * num1;
    long long sum2 =
        rem2 * (num2 + 1) * (num2 + 1) + (b + 2 - rem2) * num2 * num2;
    pq.push(make_tuple(sum1 - sum2, b + 1, c));
    cnt++;
  }
  long long ans = 0;
  while (!pq.empty()) {
    auto var = pq.top();
    pq.pop();
    long long b = get<1>(var);
    long long c = get<2>(var);
    long long num1 = arr[c] / (b);
    long long rem1 = arr[c] % (b);
    long long sum1 = rem1 * (num1 + 1) * (num1 + 1) + (b - rem1) * num1 * num1;
    ans = ans + sum1;
  }
  cout << ans;
  return 0;
}
