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 read() {
  char ch = getchar();
  int x = 0, pd = 0;
  while (ch < '0' || ch > '9') pd ^= ch == '-', ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
  return pd ? -x : x;
}
int n, m;
long long calc(long long num, int k) {
  long long t = num / k;
  return t * t * (k - num % k) + (t + 1) * (t + 1) * (num % k);
}
struct Node {
  long long num;
  int k;
  Node(long long x = 0, int y = 0) : num(x), k(y) {}
  bool operator<(Node x) const {
    return calc(num, k) - calc(num, k + 1) <
           calc(x.num, x.k) - calc(x.num, x.k + 1);
  }
};
priority_queue<Node> q;
long long ans;
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) {
    int x = read();
    q.push(Node(x, 1));
  }
  for (int i = n + 1; i <= m; i++) {
    Node t = q.top();
    q.pop(), ++t.k;
    q.push(t);
  }
  while (!q.empty()) ans += calc(q.top().num, q.top().k), q.pop();
  printf("%lld\n", ans);
  return 0;
}
