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;
static long long values[100001];
static int cs[100001];
static long long cc[100001];
long long cost_to_split(int i) {
  long long n = cs[i] + 1;
  long long x = values[i] / (cs[i] + 1);
  long long c = values[i] % n;
  return c * (2 * x + 1) + n * x * x - cc[i];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  std::vector<std::pair<int, int>> booms;
  for (int i = 0; i < n; ++i) {
    cin >> values[i];
    cc[i] = values[i] * values[i];
    cs[i] = 1;
  }
  using pli = pair<long long, int>;
  std::priority_queue<pli, vector<pli>, greater<pli>> bonuses;
  for (int i = 0; i < n; ++i) {
    bonuses.emplace(cost_to_split(i), i);
  }
  for (int i = n; i < k; ++i) {
    pli split = bonuses.top();
    bonuses.pop();
    int j = split.second;
    cc[j] += split.first;
    ++cs[j];
    bonuses.emplace(cost_to_split(j), j);
  }
  long long res = 0;
  for (int i = 0; i < n; ++i) {
    res += cc[i];
  }
  cout << res << endl;
  return 0;
}
