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 n, k;
long long cost(long long r, long long s) {
  long long a1 = r / s;
  long long r1 = r % s;
  return (s - r1) * a1 * a1 + (r1) * (a1 + 1) * (a1 + 1);
}
void solve(long long a[]) {
  priority_queue<pair<long long, long long>> p;
  vector<int> b(n, 1);
  long long c = 0, num = k - n;
  for (int i = 0; i < n; i++) c += a[i] * a[i];
  for (int i = 0; i < n; i++) {
    long long ci = cost(a[i], 1) - cost(a[i], 2);
    p.push({ci, i});
  }
  for (int i = 0; i < num; i++) {
    pair<long long, long long> tp = p.top();
    p.pop();
    c -= tp.first;
    b[tp.second]++;
    long long ci =
        cost(a[tp.second], b[tp.second]) - cost(a[tp.second], b[tp.second] + 1);
    p.push({ci, tp.second});
  }
  cout << c << endl;
}
int main() {
  cin >> n >> k;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  solve(a);
}
