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;
priority_queue<pair<long long, pair<int, int> > > heap;
long long calc(int x, int y) {
  y++;
  int ai = x / y, bi = x / y + 1, aj, bj;
  bj = (x % y);
  aj = y - bj;
  return 1ll * ai * ai * aj + 1ll * bi * bi * bj;
}
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  m -= n;
  int x;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    heap.push(make_pair(calc(x, 0) - calc(x, 1), make_pair(x, 1)));
    ans += 1ll * x * x;
  }
  pair<long long, pair<int, int> > tmp;
  int tot;
  long long val;
  while (m--) {
    tmp = heap.top();
    heap.pop();
    x = tmp.second.first;
    tot = tmp.second.second;
    ans -= tmp.first;
    val = calc(x, tot) - calc(x, tot + 1);
    heap.push(make_pair(val, make_pair(x, tot + 1)));
  }
  cout << ans << endl;
  return 0;
}
