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 cnt[100005];
int n, k;
long long cost(long long amt, long long divs) {
  divs++;
  return ((amt / divs) + 1LL) * ((amt / divs) + 1LL) * (amt % divs) +
         (amt / divs) * (amt / divs) * (divs - (amt % divs));
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%lld", a + i);
  priority_queue<pair<long long, int> > pq;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += a[i] * a[i];
    if (a[i] == 1) continue;
    pq.push(make_pair(a[i] * a[i] - cost(a[i], 1), i));
  }
  for (int i = 1; i <= k - n; i++) {
    pair<long long, int> entry = pq.top();
    pq.pop();
    ans -= entry.first;
    cnt[entry.second]++;
    if (cnt[entry.second] < a[entry.second] - 1) {
      pq.push(make_pair(cost(a[entry.second], cnt[entry.second]) -
                            cost(a[entry.second], cnt[entry.second] + 1),
                        entry.second));
    }
  }
  printf("%lld", ans);
  return 0;
}
