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 get(long long num, long long cnt) {
  long long val = num / cnt;
  long long rem = num % cnt;
  long long ans = ((val + 1) * (val + 1) * rem);
  ans = ans + ((cnt - rem) * val * val);
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  long long ans = 0;
  multiset<pair<long long, pair<long long, long long>>> Set;
  for (long long i = 0; i < n; ++i) {
    long long elem;
    cin >> elem;
    ans = ans + (elem * elem);
    long long val = get(elem, 2) - get(elem, 1);
    Set.insert({val, {elem, 2}});
  }
  long long cnt = k - n;
  while (cnt--) {
    auto it = Set.begin();
    ans = ans + ((*it).first);
    long long elem = ((*it).second.first);
    long long kk = ((*it).second.second);
    long long val = get(elem, kk + 1) - get(elem, kk);
    Set.erase(it);
    Set.insert({val, {elem, kk + 1}});
  }
  cout << ans;
  return 0;
}
