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 i, j, n, k, t;
long long x, tmp, res;
struct sb {
  long long a, b, c;
} p, p1;
bool operator<(const sb& aa, const sb& bb) { return aa.c < bb.c; }
priority_queue<sb> q;
inline long long f(long long a, long long b) {
  long long x = a / b, y = a % b;
  return x * x * (b - y) + (x + 1) * (x + 1) * y;
}
int main() {
  scanf("%d%d", &n, &k);
  k -= n;
  for (i = 1; i <= n; i++) {
    scanf("%lld", &x);
    res += x * x;
    p.a = x;
    p.b = 1;
    p.c = f(p.a, 1) - f(p.a, 2);
    q.push(p);
  }
  while (k--) {
    p = q.top();
    q.pop();
    res -= p.c;
    p.b++;
    p.c = f(p.a, p.b) - f(p.a, p.b + 1);
    q.push(p);
  }
  printf("%lld\n", res);
}
