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;
void fff(string s = "") {
  if (s == "1") cin >> s;
  if ((int)s.size() >= 4 && s.substr(s.size() - 4) == ".inp") {
    freopen(s.c_str(), "r", stdin);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  fff("");
  int n, k;
  cin >> n >> k;
  long long ans = 0;
  priority_queue<pair<long long, int>> pq;
  auto get = [&](int val, int num) {
    int x = val / num;
    int odd = val % num;
    return 1ll * (num - odd) * x * x + 1ll * odd * (x + 1) * (x + 1);
  };
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ans += get(a[i], 1);
    pq.emplace(get(a[i], 1) - get(a[i], 2), i);
  }
  k -= n;
  vector<int> cnt(n, 1);
  while (k--) {
    auto x = pq.top();
    pq.pop();
    ans -= x.first;
    cnt[x.second] += 1;
    pq.emplace(
        get(a[x.second], cnt[x.second]) - get(a[x.second], cnt[x.second] + 1),
        x.second);
  }
  cout << ans;
}
