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;
inline int64_t value(int len, int p) {
  if (len <= p) return len;
  int64_t q = len / p;
  int r = len % p;
  return q * q * (p - r) + (q + 1) * (q + 1) * r;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  int64_t cost = 0;
  priority_queue<tuple<int64_t, int64_t, int, int>> pq;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    int64_t c = (int64_t)a * a;
    int64_t nxt = value(a, 2);
    cost += c;
    if (nxt < c) pq.emplace(c - nxt, nxt, a, 2);
  }
  for (int x = n; x < k; x++) {
    auto [gained, now, len, pieces] = pq.top();
    pq.pop();
    cost -= gained;
    int64_t nxt = value(len, pieces + 1);
    if (nxt < now) pq.emplace(now - nxt, nxt, len, pieces + 1);
  }
  cout << cost << '\n';
  return 0;
}
