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;
constexpr static int MAXN = 1e5 + 10;
int n, k;
int a[MAXN];
int num[MAXN];
int64_t calc(int64_t val, int m) {
  int64_t base = val / m;
  int64_t rem = val % m;
  if (rem == 0) return base * base * m;
  return base * base * (m - rem) + (base + 1) * (base + 1) * rem;
}
int64_t calc_diff(int64_t val, int m) {
  return calc(val, m) - calc(val, m + 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  priority_queue<pair<int64_t, int>> pq;
  for (int i = 0; i < n; i++) {
    num[i] = 1;
    if (a[i] == 1) continue;
    pq.push({calc_diff(a[i], 1), i});
  }
  for (int l = 0; l < k - n; l++) {
    auto t = pq.top();
    pq.pop();
    int i = t.second;
    num[i]++;
    if (num[i] < a[i]) pq.push({calc_diff(a[i], num[i]), i});
  }
  int64_t total = 0;
  for (int i = 0; i < n; i++) total += calc(a[i], num[i]);
  cout << total << endl;
}
