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 S(long long l, long long k) {
  long long len1 = l / k;
  long long len2 = len1 + 1;
  long long cnt2 = l % k;
  long long cnt1 = k - cnt2;
  return cnt1 * len1 * len1 + cnt2 * len2 * len2;
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<long long> v(n);
  long long ans = 0;
  priority_queue<pair<long long, pair<long long, long long>>> q;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    ans += v[i] * v[i];
    q.push({S(v[i], 1) - S(v[i], 2), {v[i], 2}});
  }
  for (int i = 0; i < k - n; i++) {
    auto t = q.top();
    q.pop();
    ans -= t.first;
    int a = t.second.first, b = t.second.second;
    q.push({S(a, b) - S(a, b + 1), {a, b + 1}});
  }
  cout << ans;
  return 0;
}
