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 long long MAXN = 3e5 + 5;
const long long mod = 1e9 + 7;
long long b[MAXN], a[MAXN];
long long get(long long x, long long k) {
  if (a[x] < k) return 1e9;
  long long duan = a[x] / k;
  long long num = a[x] % k;
  return num * (duan + 1) * (duan + 1) + (k - num) * (duan) * (duan);
}
struct node {
  long long x, y;
  node() {}
  node(long long a, long long b) { x = a, y = b; }
  bool operator<(const node& r) const { return y < r.y; }
};
int main() {
  long long n, k;
  cin >> n >> k;
  priority_queue<node> q;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = 1;
    q.push(node(i, get(i, 1) - get(i, 2)));
  }
  long long num = k - n;
  while (num--) {
    node x = q.top();
    q.pop();
    ++b[x.x];
    long long now = get(x.x, b[x.x]) - get(x.x, b[x.x] + 1);
    q.push(node(x.x, now));
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    ans += get(i, b[i]);
  }
  cout << ans << endl;
}
