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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k, ans = 0;
  cin >> n >> k;
  vector<long long> a(n), b(n), cnt(n, 2);
  const long long inf = 1e18;
  set<pair<long long, int>, greater<pair<long long, int>>> s;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i] * a[i];
    ans += b[i];
    long long t = a[i] / 2;
    long long better = t * t + (a[i] - t) * (a[i] - t);
    s.insert(make_pair(b[i] - better, i));
  }
  for (int i = 0; i < k - n; i++) {
    auto p = *s.begin();
    s.erase(s.begin());
    ans -= p.first;
    cnt[p.second]++;
    b[p.second] -= p.first;
    long long temp = a[p.second] / cnt[p.second];
    long long q = a[p.second] % cnt[p.second];
    long long better =
        temp * temp * (cnt[p.second] - q) + (temp + 1) * (temp + 1) * q;
    s.insert(make_pair(b[p.second] - better, p.second));
  }
  cout << ans << endl;
  return 0;
}
