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 ar[200005];
char ch[200005];
int n, m, ii, k;
long long Cost(long long x, int part) {
  long long per = x / part;
  long long boro = x % part;
  long long choto = part - boro;
  return choto * per * per + boro * (per + 1) * (per + 1);
}
void solve() {
  scanf("%d%d", &n, &k);
  vector<int> Cut(n + 5, 1);
  priority_queue<pair<long long, int>> pq;
  long long re = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ar[i]);
    re += Cost(ar[i], 1);
    if (ar[i] > Cut[i])
      pq.push({Cost(ar[i], Cut[i]) - Cost(ar[i], Cut[i] + 1), i});
  }
  while (k > n) {
    k--;
    re -= pq.top().first;
    int i = pq.top().second;
    pq.pop();
    Cut[i]++;
    if (ar[i] > Cut[i])
      pq.push({Cost(ar[i], Cut[i]) - Cost(ar[i], Cut[i] + 1), i});
  }
  printf("%lld\n", re);
}
int main() {
  int t = 1;
  while (t--) solve();
  return 0;
}
