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;
int n, k;
vector<int> a;
struct info {
  int id, split;
  info(int _id, int _split) {
    id = _id;
    split = _split;
  }
  long long sqr(int a) { return 1LL * a * a; }
  long long cost() {
    int x = a[id] / split;
    int y = a[id] % split;
    return (split - y) * sqr(x) + y * sqr(x + 1);
  }
  bool operator<(const info &A) const {
    long long p1 = info(id, split).cost() - info(id, split + 1).cost();
    long long p2 = info(A.id, A.split).cost() - info(A.id, A.split + 1).cost();
    return p1 < p2;
  }
};
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (fopen("main.in", "r")) freopen("main.in", "r", stdin);
  cin >> n >> k;
  a.resize(n + 1);
  for (int i = 1; i <= n; i++) cin >> a[i];
  long long res = 0;
  priority_queue<info> heap;
  for (int i = 1; i <= n; i++) {
    heap.push({i, 1});
    res += info(i, 1).cost();
  }
  for (int i = n + 1; i <= k; i++) {
    info it = heap.top();
    heap.pop();
    info nxt = {it.id, it.split + 1};
    res -= it.cost() - nxt.cost();
    heap.push(nxt);
  }
  cout << res << '\n';
  return 0;
}
