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;
template <typename T>
using vc = vector<T>;
using pii = pair<long long, long long>;
void xmax(long long& a, long long b) { a = max(a, b); }
void xmin(long long& a, long long b) { a = min(a, b); }
void print(vc<long long>& a, string second) {
  cerr << second << " : ";
  for (long long i = 0; i < a.size(); i++) {
    cerr << a[i] << " ";
  }
  cerr << endl;
}
long long find_anz(long long gr, long long anz) {
  if (anz > gr) return -3;
  long long p = gr / anz;
  long long anz_p = anz - gr % anz;
  long long plu = p + 1;
  long long anz_plu = gr % anz;
  long long res = p * p * anz_p + plu * plu * anz_plu;
  return res;
}
void do_test() {
  long long n, k;
  cin >> n >> k;
  vc<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vc<long long> b(n, 1);
  long long tot = 0;
  for (long long i = 0; i < n; i++) {
    tot += a[i] * a[i];
  }
  priority_queue<pii> pq;
  for (long long i = 0; i < n; i++) {
    if (a[i] != 1) pq.push({find_anz(a[i], 1) - find_anz(a[i], 2), i});
  }
  for (long long i = n; i < k; i++) {
    auto tap = pq.top();
    pq.pop();
    assert(tap.first > 0);
    tot -= tap.first;
    long long ind = tap.second;
    b[ind]++;
    if (b[ind] < a[ind])
      pq.push({find_anz(a[ind], b[ind]) - find_anz(a[ind], b[ind] + 1), ind});
  }
  cout << tot << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  do_test();
}
