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>
#pragma GCC optimize("Ofast")
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << ", ";
  err(++it, args...);
  cout << "\n";
}
long long powm(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
const long long mod = 1e9 + 7;
const long long N = 1000005;
const long long inf = 1e9;
vector<long long> a;
long long calc(long long len, long long nums) {
  long long unit = len / nums;
  long long extra = len - unit * nums;
  return (nums - extra) * (unit * unit) + extra * (unit + 1) * (unit + 1);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long x;
  priority_queue<pair<long long, pair<long long, long long>>> v;
  long long sum = 0;
  for (long long i = 0; i < (n); i++) {
    cin >> x;
    sum += calc(x, 1);
    v.push({calc(x, 1) - calc(x, 2), {x, 2}});
  }
  for (long long i = 0; i < (k - n); i++) {
    auto it = v.top();
    sum -= it.first;
    v.pop();
    v.push({calc(it.second.first, it.second.second) -
                calc(it.second.first, it.second.second + 1),
            {it.second.first, it.second.second + 1}});
  }
  cout << sum << '\n';
}
signed main() {
  cout << fixed << setprecision(15);
  int tc = 1;
  for (int t = 1; t <= tc; t++) {
    solve();
  }
  return 0;
}
