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 = 1e5;
int a[MaxN + 5], n, k;
long long ans = 0LL;
long long getval(int x, int y) {
  int cnt = x / y;
  int m = x % y;
  return 1LL * m * (cnt + 1) * (cnt + 1) + 1LL * (y - m) * cnt * cnt;
}
struct node {
  long long val;
  int x, block;
  bool operator<(const node &a) const { return val < a.val; }
};
priority_queue<node> que;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    ans += 1LL * a[i] * a[i];
    que.push((node){getval(a[i], 1) - getval(a[i], 2), a[i], 1});
  }
  int cnt = k - n;
  while (cnt) {
    --cnt;
    node now = que.top();
    que.pop();
    ans -= now.val;
    if (now.block == now.x) continue;
    que.push((node){getval(now.x, now.block + 1) - getval(now.x, now.block + 2),
                    now.x, now.block + 1});
  }
  cout << ans << "\n";
  return 0;
}
