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 MAXN = 2e5 + 5;
int n, k;
int a[MAXN];
long long totAns = 0;
long long getRedux(int init, int cut) {
  long long initVal = 1LL * init * init;
  long long newVal = 0;
  long long tmp = init / (cut);
  newVal = ((tmp + 1) * (tmp + 1)) * (init % cut);
  newVal += (tmp * tmp) * (cut - (init % cut));
  return newVal;
}
int main() {
  priority_queue<pair<long long, pair<int, int> > > q;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    totAns += 1LL * a[i] * a[i];
    if (a[i] > 1)
      q.push(pair<long long, pair<int, int> >(
          (1LL * a[i] * a[i]) - getRedux(a[i], 2), pair<int, int>(a[i], 2)));
  }
  while (k > n) {
    k--;
    pair<long long, pair<int, int> > top = q.top();
    q.pop();
    totAns -= top.first;
    pair<int, int> conf = top.second;
    if (conf.second < conf.first) {
      long long newRedux = getRedux(conf.first, conf.second) -
                           getRedux(conf.first, conf.second + 1);
      q.push(pair<long long, pair<int, int> >(
          newRedux, pair<int, int>(conf.first, conf.second + 1)));
    }
  }
  printf("%lld\n", totAns);
}
