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 elem {
  long long x, c = 1;
  long long val;
  long long nval() {
    long long r = x % (c + 1);
    return x / (c + 1) * (x / (c + 1) * (c + 1 - r)) +
           (x / (c + 1) + 1) * ((x / (c + 1) + 1) * r);
  }
  friend bool operator<(elem a, elem b) {
    long long ca = a.val - a.nval(), cb = b.val - b.nval();
    return ca < cb;
  }
};
long long n, k;
int32_t main() {
  cin >> n >> k;
  k = k - n;
  long long sol = 0;
  priority_queue<elem> q;
  for (long long i = 1; i <= n; ++i) {
    elem x;
    cin >> x.x;
    x.c = 1;
    x.val = x.x * x.x;
    sol += x.val;
    q.push(x);
  }
  while (k--) {
    elem x = q.top();
    q.pop();
    sol -= x.val;
    x.val = x.nval();
    x.c++;
    sol += x.val;
    q.push(x);
  }
  cout << sol << "\n";
}
