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;
void PrintArrInt(int arr[], int s, int e) {
  for (int i = s; i <= e; i++) {
    if (i == s)
      printf("%d", arr[i]);
    else
      printf(" %d", arr[i]);
  }
  putchar('\n');
}
const int maxn = 100000 + 5;
int N, K, arr[maxn];
struct Node {
  long long d;
  int val, div;
  bool operator<(const Node& rhs) const { return d < rhs.d; }
};
priority_queue<Node> q;
long long calc(int a, int div) {
  long long t = a / div, tm = a % div;
  return tm * (t + 1) * (t + 1) + (div - tm) * t * t;
}
int main() {
  scanf("%d%d", &(N), &(K));
  for (int i = (1); i <= (N); i++) scanf("%d", &(arr[i]));
  long long sum = 0;
  for (int i = (1); i <= (N); i++) {
    long long d = 1ll * arr[i] * arr[i];
    sum += d;
    d -= calc(arr[i], 2);
    q.push(Node{d, arr[i], 2});
  }
  K -= N;
  while (K--) {
    Node n = q.top();
    q.pop();
    sum -= n.d;
    q.push(Node{calc(n.val, n.div) - calc(n.val, n.div + 1), n.val, n.div + 1});
  }
  printf("%lld\n", (sum));
  return 0;
}
