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;
long long arr[1100000];
long long f(int l, int p) {
  long long w = l / p;
  long long p2 = l % p;
  return (p - p2) * w * w + p2 * (w + 1) * (w + 1);
}
struct cmp {
  bool operator()(const pair<long long, long long>& a,
                  const pair<long long, long long>& b) {
    long long aa = f(a.first, a.second) - f(a.first, a.second + 1);
    long long bb = f(b.first, b.second) - f(b.first, b.second + 1);
    if (aa == bb)
      return a.first < b.first;
    else
      return aa < bb;
  }
};
void solve(int tc) {
  cin >> N >> K;
  for (int i = 0; i < N; i++) cin >> arr[i];
  priority_queue<pair<long long, long long>,
                 vector<pair<long long, long long> >, cmp>
      pq;
  for (int i = 0; i < N; i++) pq.push({arr[i], 1});
  for (int i = 0; i < K - N; i++) {
    pair<long long, long long> cur = pq.top();
    pq.pop();
    pq.push({cur.first, cur.second + 1});
  }
  long long ans = 0;
  while (pq.size()) {
    ans += f(pq.top().first, pq.top().second);
    pq.pop();
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  if (0) {
    int T;
    cin >> T;
    for (int i = 1; i <= T; i++) solve(i);
  } else {
    solve(0);
  }
  return 0;
}
