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;
long long f(long long k, long long p) {
  long long fl = k / p;
  long long b = k % p, a = p - b;
  return (a * (fl * fl)) + b * (fl + 1) * (fl + 1);
}
priority_queue<pair<long long, pair<long long, long long>>> pq;
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    long long nextStepLav = f(x, 1) - f(x, 2);
    pq.push({nextStepLav, {x, 1}});
  }
  for (int i = 1; i <= k - n; i++) {
    auto u = pq.top();
    pq.pop();
    long long k = u.second.first;
    long long p = u.second.second;
    long long nextStepLav = f(k, p + 1) - f(k, p + 2);
    u = {nextStepLav, {k, p + 1}};
    pq.push(u);
  }
  long long ans = 0;
  while (!pq.empty()) {
    auto u = pq.top();
    pq.pop();
    long long k = u.second.first;
    long long p = u.second.second;
    ans += f(k, p);
  }
  cout << ans;
}
