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;
const int N = 1e4 + 7;
const int lim = 1000;
const int MOD = 1e9 + 7;
const double eps = 1e-6;
struct Node {
  long long a, cs, sum, cha;
  bool operator<(const Node &b) const { return cha < b.cha; }
};
priority_queue<Node> pq;
int n, k;
long long x;
long long fun(long long a, long long b) {
  long long x = a / b;
  long long y = a % b;
  long long ans = x * x * (b - y) + (x + 1) * (x + 1) * y;
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &x);
    Node tmp;
    tmp.a = x;
    tmp.sum = x * x;
    tmp.cs = 1;
    tmp.cha = fun(x, 1) - fun(x, 2);
    pq.push(tmp);
  }
  k -= n;
  while (k > 0) {
    Node tmp = pq.top();
    pq.pop();
    Node kk;
    kk.a = tmp.a;
    kk.cs = tmp.cs + 1;
    kk.sum = fun(kk.a, kk.cs);
    kk.cha = fun(kk.a, kk.cs) - fun(kk.a, kk.cs + 1);
    pq.push(kk);
    k--;
  }
  long long ans = 0;
  while (!pq.empty()) {
    Node tmp = pq.top();
    pq.pop();
    ans += tmp.sum;
  }
  printf("%lld\n", ans);
  return 0;
}
