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;
const int N = 1e5;
struct car {
  int a, s;
  long long f, g;
  long long d(const int &k) const {
    const int c = a / k;
    return 1ll * k * c * c + 1ll * (a % k) * (2 * c + 1);
  }
  car(int a) : a(a) { f = d(s = 1), g = d(2); }
  void split() { f = g, g = d(1 + ++s); }
  friend bool operator<(const car &p, const car &q) {
    return p.g - p.f > q.g - q.f;
  }
};
priority_queue<car> q;
int n, k;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  long long res = 0;
  for (int i = (0); i < (n); ++i) {
    int a;
    cin >> a;
    car c = a;
    res += c.f, q.push(c);
  }
  for (int t = k - n; t; --t) {
    car c = q.top();
    q.pop();
    res += c.g - c.f, c.split(), q.push(c);
  }
  cout << res << '\n';
  return 0;
}
