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;
struct Node {
  long long a, b, c;
  Node(long long a_ = 0, long long b_ = 0, long long c_ = 0)
      : a(a_), b(b_), c(c_) {}
  bool operator<(const Node &B) const { return c < B.c; }
};
priority_queue<Node> Q;
long long deal(long long a, long long b) {
  long long c = a / b, d = a % b;
  return d * (c + 1) * (c + 1) + (b - d) * c * c;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    ans += (long long)a * a;
    Q.emplace(a, 1, deal(a, 1) - deal(a, 2));
  }
  for (int i = 1; i <= k - n; i++) {
    Node u = Q.top();
    Q.pop();
    ans -= u.c;
    Q.emplace(u.a, u.b + 1, deal(u.a, u.b + 1) - deal(u.a, u.b + 2));
  }
  cout << ans;
  return 0;
}
