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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  k -= n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  auto get = [&](int x, int y) {
    int l = x / y;
    int h = l + 1;
    int cnth = x % y;
    int cntl = y - cnth;
    return (long long)l * l * cntl + (long long)h * h * cnth;
  };
  long long ans = 0;
  set<pair<long long, int>> have;
  vector<int> num(n, 1);
  for (int i = 0; i < n; ++i) {
    ans += (long long)a[i] * a[i];
    if (a[i] > num[i]) {
      have.insert(make_pair(-get(a[i], num[i]) + get(a[i], num[i] + 1), i));
    }
  }
  for (int tt = 0; tt < k; ++tt) {
    auto p = *have.begin();
    have.erase(have.begin());
    ans += p.first;
    int i = p.second;
    ++num[i];
    if (a[i] > num[i]) {
      have.insert(make_pair(-get(a[i], num[i]) + get(a[i], num[i] + 1), i));
    }
  }
  cout << ans << '\n';
}
