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 maxn = 1e6 + 6;
struct node {
  long long x, y, d;
  bool operator<(const node &a) const { return d < a.d; }
};
long long inv(long long a, long long b) {
  long long t = a / b;
  long long mod = a % b;
  return t * t * (b - mod) + mod * (t + 1) * (t + 1);
}
int main() {
  long long n, k, a;
  while (cin >> n >> k) {
    priority_queue<node> qu;
    long long ans = 0;
    for (long long i = 1; i <= n; i++) {
      cin >> a;
      qu.push(node{a, 1, inv(a, 1) - inv(a, 2)});
      ans += a * a;
    }
    while (n < k) {
      ++n;
      node top = qu.top();
      qu.pop();
      ans -= top.d;
      qu.push(node{top.x, top.y + 1,
                   inv(top.x, top.y + 1) - inv(top.x, top.y + 2)});
    }
    cout << ans << endl;
  }
  return 0;
}
