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 a[100005];
int par[100005];
priority_queue<pair<long long, int> > pq;
long long tim(long long carrot, int parts) {
  long long w = carrot / parts;
  return (w) * (w) * (parts - carrot % parts) +
         (w + 1) * (w + 1) * (carrot % parts);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    par[i] = 1;
    pq.push({tim(a[i], 1) - tim(a[i], 2), i});
  }
  for (int i = 0; i < k - n; i++) {
    pair<long long, int> top = pq.top();
    pq.pop();
    long long carrot = a[top.second];
    int parts = ++par[top.second];
    pq.push({tim(carrot, parts) - tim(carrot, parts + 1), top.second});
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += tim(a[i], par[i]);
  }
  cout << ans << endl;
  return 0;
}
