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 maxn = 2e5 + 5;
int a[maxn];
long long cal(int x, int num) {
  int p = x / num, q = x % num;
  return 1ll * (p + 1) * (p + 1) * q + 1ll * p * p * (num - q);
}
struct node {
  int v, cnt;
  bool operator<(const node &a) const {
    return cal(v, cnt) - cal(v, cnt + 1) <
           cal(a.v, a.cnt) - cal(a.v, a.cnt + 1);
  }
};
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  priority_queue<node> q;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    ans += 1ll * a[i] * a[i];
    q.push(node{a[i], 1});
  }
  for (int i = n + 1; i <= k; i++) {
    node now = q.top();
    q.pop();
    ans -= cal(now.v, now.cnt) - cal(now.v, now.cnt + 1);
    q.push(node{now.v, now.cnt + 1});
  }
  printf("%lld\n", ans);
}
