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;
using ll = long long;
ll Square(ll x) { return x * x; }
ll Divide(ll x, ll d) {
  d = min(d, x);
  ll group1 = d - x % d;
  ll group2 = x % d;
  ll res = group1 * Square(x / d) + group2 * Square(x / d + 1);
  return res;
}
int main() {
  ll i, j, k;
  ll n;
  cin >> n >> k;
  priority_queue<pair<ll, pair<ll, ll>>> pq;
  ll res = 0;
  for (i = 0; i < n; ++i) {
    cin >> j;
    res += Divide(j, 1);
    pq.push({Divide(j, 1) - Divide(j, 2), {j, 1}});
  }
  for (i = n; i < k; ++i) {
    pair<ll, ll> p = pq.top().second;
    res -= Divide(p.first, p.second);
    res += Divide(p.first, p.second + 1);
    pq.pop();
    pq.push({Divide(p.first, p.second + 1) - Divide(p.first, p.second + 2),
             {p.first, p.second + 1}});
  }
  cout << res << endl;
}
