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 sqr(int x) { return 1ll * x * x; }
using ll = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  vector<int> a(n), block(n, 0);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long ans = 0;
  priority_queue<pair<ll, int>, vector<pair<ll, int> >,
                 greater<pair<ll, int> > >
      q;
  auto get = [&](int i, int piece) {
    int temp = a[i] % piece;
    long long res = 0;
    res += 1ll * sqr(a[i] / piece) * (piece - temp);
    res += 1ll * sqr(a[i] / piece + 1) * temp;
    return res;
  };
  for (int i = 0; i < n; i++) {
    ans += 1ll * a[i] * a[i];
    ++block[i];
    q.emplace(get(i, 2) - get(i, 1), i);
    --k;
  }
  while (k--) {
    auto u = q.top();
    q.pop();
    ans += u.first;
    auto i = u.second;
    ++block[i];
    q.emplace(get(i, block[i] + 1) - get(i, block[i]), i);
  }
  cout << ans << '\n';
  return 0;
}
