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;
struct qq {
  long long a, b, c;
  bool operator<(const qq &b) const { return a < b.a; }
};
priority_queue<qq> Q;
long long js(long long a, long long x) {
  long long m = a / x, t = a % x;
  return t * (m + 1) * (m + 1) + (x - t) * m * m;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, k;
  cin >> n >> k;
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    qq t;
    t.a = js(x, 1) - js(x, 2);
    t.b = x;
    t.c = 2;
    Q.push(t);
    sum += x * x;
  }
  for (int i = 1; i <= k - n; i++) {
    qq x = Q.top();
    Q.pop();
    sum -= x.a;
    Q.push({js(x.b, x.c) - js(x.b, x.c + 1), x.b, x.c + 1});
  }
  cout << sum << "\n";
  return 0;
}
