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 = 2e5 + 10;
int n, k;
long long a[maxn];
priority_queue<pair<long long, pair<int, long long> >,
               vector<pair<long long, pair<int, long long> > >,
               less<pair<long long, pair<int, long long> > > >
    q;
long long cal(long long first, int c) {
  long long f = first / c;
  long long now =
      1LL * (c - first % c) * f * f + (first % c) * (f + 1) * (f + 1);
  c++;
  f = first / c;
  long long last =
      1LL * (c - first % c) * f * f + (first % c) * (f + 1) * (f + 1);
  return now - last;
}
int main() {
  scanf("%d%d", &n, &k);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    q.push({cal(a[i], 1), {1, a[i]}});
    ans += a[i] * a[i];
  }
  int cnt = k - n;
  for (pair<long long, pair<int, long long> > to = q.top(); cnt; to = q.top()) {
    q.pop();
    ans -= to.first;
    q.push({cal(to.second.second, to.second.first + 1),
            {to.second.first + 1, to.second.second}});
    cnt--;
  }
  printf("%lld\n", ans);
}
