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(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  long long a[n];
  long long res = 0;
  priority_queue<pair<long long, pair<int, int>>> pq;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    res += a[i] * a[i];
    if (a[i] > 1) {
      long long x = a[i] / 2;
      long long y = x + (a[i] & 1);
      long long diff = a[i] * a[i] - y * y - x * x;
      pq.push({diff, {i, 2}});
    }
  }
  int cnt = n;
  while (cnt < k) {
    int id = pq.top().second.first;
    int div = pq.top().second.second + 1;
    res -= pq.top().first;
    pq.pop();
    cnt++;
    if (div > a[id]) continue;
    long long xp, yp, rp, xq, yq, rq, p, q;
    xp = a[id] / div;
    yp = xp + 1;
    rp = a[id] % div;
    p = (div - rp) * xp * xp + rp * yp * yp;
    xq = a[id] / (div - 1);
    yq = xq + 1;
    rq = a[id] % (div - 1);
    q = (div - 1 - rq) * xq * xq + rq * yq * yq;
    pq.push({q - p, {id, div}});
  }
  cout << res << '\n';
}
