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;
const int N = 2e5 + 7;
long long a[N];
long long ccal(int n, int k) {
  long long aa = n / k;
  long long b = n % k;
  long long kk = k - b;
  return 1ll * kk * (aa * aa) + 1ll * b * (aa + 1) * (aa + 1);
}
long long cal(int n, int k) { return ccal(n, k - 1) - ccal(n, k); }
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += a[i] * a[i];
  }
  priority_queue<pair<long long, pair<int, int> > > q;
  for (int i = 1; i <= n; i++) {
    q.push(make_pair(cal(a[i], 2), make_pair(a[i], 2)));
  }
  int res = k - n;
  while (res--) {
    auto x = q.top();
    ans -= x.first;
    q.pop();
    q.push(make_pair(cal(x.second.first, x.second.second + 1),
                     make_pair(x.second.first, x.second.second + 1)));
  }
  cout << ans << '\n';
  ;
  return 0;
}
