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 inf = 10e5;
long long summ(long long initial_len, long long cuts) {
  if (initial_len % cuts == 0) {
    return initial_len * initial_len / cuts;
  } else {
    long long rem = initial_len % cuts;
    long long each = initial_len / cuts;
    return (each + 1) * (each + 1) * rem + each * each * (cuts - rem);
  }
}
long long profit(long long initial_len, long long cuts) {
  return summ(initial_len, cuts) - summ(initial_len, cuts + 1);
}
int main() {
  long long n, k;
  cin >> n >> k;
  long long a[n];
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    ans += a[i] * a[i];
  }
  long long times = k - n;
  priority_queue<tuple<long long, long long, long long>> q;
  for (long long i = 0; i < n; i++) {
    q.push(make_tuple(profit(a[i], 1), a[i], 1));
  }
  for (long long i = 0; i < times; i++) {
    long long prof = get<0>(q.top());
    long long val = get<1>(q.top());
    long long curr_cuts = get<2>(q.top());
    q.pop();
    ans -= prof;
    q.push(make_tuple(profit(val, curr_cuts + 1), val, curr_cuts + 1));
  }
  cout << ans << endl;
  return 0;
}
