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 N = 2e5 + 5;
const int mod = 998244353;
long long get(int x, int slice) {
  int sz = x / slice;
  int m = x % slice;
  long long ret =
      (long long)sz * sz * (slice - m) + (long long)(sz + 1) * (sz + 1) * m;
  return ret;
}
void solve() {
  int n, k;
  scanf("%d %d", &n, &k);
  priority_queue<pair<long long, pair<int, int> > > pq;
  int cur = n;
  long long ret = 0;
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    ret += (long long)x * x;
    if (x != 1) {
      pq.push({get(x, 1) - get(x, 2), {x, 1}});
    }
  }
  while (cur < k) {
    ret -= pq.top().first;
    pair<int, int> t = pq.top().second;
    pq.pop();
    ++cur;
    ++t.second;
    if (t.first >= t.second + 1) {
      pq.push({get(t.first, t.second) - get(t.first, t.second + 1),
               {t.first, t.second}});
    }
  }
  printf("%lld\n", ret);
}
int main() {
  solve();
  return 0;
}
