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 long long N = 1e5 + 5;
long long n, k;
long long a[N];
long long get(long long cur, long long divs) {
  long long sum = 0;
  long long val = cur / divs;
  long long extra = cur % divs;
  for (long long j = 1; j <= divs; j++) {
    if (extra >= j)
      sum += (val + 1) * (val + 1);
    else
      sum += val * val;
  }
  return sum;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  multiset<pair<long long, pair<long long, long long> > > s;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    pair<long long, pair<long long, long long> > cur = {
        get(a[i], 1) - get(a[i], 2), {a[i], 1}};
    s.insert(cur);
  }
  long long cur = n;
  while (cur < k) {
    cur++;
    auto cur = s.end();
    --cur;
    auto curp = *cur;
    s.erase(cur);
    long long divs = curp.second.second;
    long long val = curp.second.first;
    divs++;
    long long nw = get(val, divs) - get(val, divs + 1);
    s.insert({nw, {val, divs}});
  }
  long long sum = 0;
  for (auto &it : s) {
    long long cur = it.second.first;
    long long divs = it.second.second;
    long long val = cur / divs;
    long long extra = cur % divs;
    for (long long j = 1; j <= divs; j++) {
      if (extra >= j)
        sum += (val + 1) * (val + 1);
      else
        sum += val * val;
    }
  }
  cout << sum;
  return 0;
}
