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;
long long Cost(int a, int k) {
  int rem = a % k;
  int dv = a / k;
  return 1LL * dv * dv * (k - rem) + 1LL * (dv + 1) * (dv + 1) * rem;
}
long long CostDiff(int a, int k) {
  if (a == k) return 0;
  return Cost(a, k) - Cost(a, k + 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int& e : a) {
    cin >> e;
  }
  priority_queue<tuple<long long, int, int>> q;
  for (int e : a) {
    q.push({CostDiff(e, 1), e, 1});
  }
  for (int i = 0; i < k - n; ++i) {
    auto item = q.top();
    q.pop();
    int a = get<1>(item);
    int k = get<2>(item);
    q.push({CostDiff(a, k + 1), a, k + 1});
  }
  long long result = 0;
  while (!q.empty()) {
    auto item = q.top();
    q.pop();
    int a = get<1>(item);
    int k = get<2>(item);
    result += Cost(a, k);
  }
  cout << result << '\n';
}
