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 data_t {
  int a, b;
  long long cost() const {
    long long ans = 0;
    int a = this->a;
    int b = this->b;
    ans -= (1LL * (a % b) * (a / b + 1) * (a / b + 1) +
            1LL * (b - (a % b)) * (a / b) * (a / b));
    b++;
    ans += (1LL * (a % b) * (a / b + 1) * (a / b + 1) +
            1LL * (b - (a % b)) * (a / b) * (a / b));
    b--;
    return ans;
  }
  bool operator<(const data_t &other) const {
    return this->cost() > other.cost();
  }
};
int main() {
  int n, k;
  cin >> n >> k;
  priority_queue<data_t> pq;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int val;
    cin >> val;
    pq.push({val, 1});
    ans += 1LL * val * val;
  }
  k -= n;
  while (k--) {
    data_t tmp = pq.top();
    pq.pop();
    ans += tmp.cost();
    tmp.b++;
    pq.push(tmp);
  }
  cout << ans;
  return 0;
}
