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;
int n, k;
long long a[100100];
priority_queue<pair<long long, int> > q;
long long ans = 0;
int cc[100100];
long long gnum(int now, int c) {
  if (c == 0) {
    return a[now] * a[now];
  }
  c++;
  long long tmp = a[now] / c;
  long long cha = a[now] % c;
  long long ret = 0;
  ret += tmp * tmp * (c - cha);
  ret += (tmp + 1) * (tmp + 1) * (cha);
  return ret;
}
void push(int now, int c) {
  long long ret = gnum(now, c - 1) - gnum(now, c);
  q.push(make_pair(ret, now));
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    ans += a[i] * a[i];
    push(i, 1);
  }
  k -= n;
  while (k--) {
    auto it = q.top();
    q.pop();
    ans -= it.first;
    cc[it.second]++;
    push(it.second, cc[it.second] + 1);
  }
  printf("%lld\n", ans);
}
