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;
long long f(int l, int p) {
  int l1 = l / p;
  int p2 = l - l1 * p;
  int p1 = p - p2;
  int l2 = l1 + 1;
  return 1LL * l1 * l1 * p1 + 1LL * l2 * l2 * p2;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> arr(n);
  for (int &it : arr) {
    cin >> it;
  }
  priority_queue<pair<long long, int>, vector<pair<long long, int>>> pq;
  vector<int> pieces(n, 1);
  for (int i = 0; i < n; i++) {
    pq.push({1LL * arr[i] * arr[i] - f(arr[i], 2), i});
  }
  int total = n;
  while (total < k) {
    auto v = pq.top();
    pq.pop();
    pieces[v.second]++;
    pq.push({f(arr[v.second], pieces[v.second]) -
                 f(arr[v.second], pieces[v.second] + 1),
             v.second});
    total++;
  }
  long long answer = 0;
  for (int i = 0; i < n; i++) {
    answer += f(arr[i], pieces[i]);
  }
  cout << answer << '\n';
}
