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 f(long long l, long long p) {
  long long p1, p2;
  p2 = l % p;
  p1 = p - p2;
  return p1 * (l / p) * (l / p) + p2 * (l / p + 1) * (l / p + 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k;
  cin >> n >> k;
  vector<long long> cnt(n, 1), a(n);
  for (auto& x : a) {
    cin >> x;
  }
  auto cmp = [&](const int& i, const int& j) {
    return f(a[i], cnt[i]) - f(a[i], cnt[i] + 1) <
           f(a[j], cnt[j]) - f(a[j], cnt[j] + 1);
  };
  priority_queue<int, vector<int>, decltype(cmp)> pq(cmp);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    pq.push(i);
    ans += f(a[i], cnt[i]);
  }
  int tot = k - n;
  while (tot--) {
    int id = pq.top();
    pq.pop();
    long long before = f(a[id], cnt[id]), after = f(a[id], ++cnt[id]);
    ans += (after - before);
    pq.push(id);
  }
  cout << ans << '\n';
  return 0;
}
