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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  ll a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  k -= n;
  int c[n];
  for (int i = 0; i < n; i++) c[i] = 1;
  auto f = [&](int i, int k) {
    if (k >= a[i])
      return a[i];
    else {
      ll x = a[i] / k;
      ll y = a[i] % k;
      return y * (x + 1) * (x + 1) + (k - y) * x * x;
    }
  };
  set<pair<ll, int>> X;
  for (int i = 0; i < n; i++) X.insert({f(i, c[i] + 1) - f(i, c[i]), i});
  for (int t = 0; t < k; t++) {
    auto it = X.begin();
    int i = it->second;
    X.erase(it);
    c[i]++;
    X.insert({f(i, c[i] + 1) - f(i, c[i]), i});
  }
  ll ret = 0;
  for (int i = 0; i < n; i++) ret += f(i, c[i]);
  cout << ret << "\n";
}
