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;
long long int a[(int)1e5];
long long int f(int i, int d) {
  if (a[i] == 1) return 0;
  long long int mn = a[i] / d;
  long long int mx = mn + !!(a[i] % d);
  long long int cur = mn * mn * (d - a[i] % d);
  cur += mx * mx * (a[i] % d);
  d++;
  mn = a[i] / d;
  mx = mn + !!(a[i] % d);
  cur -= mn * mn * (d - a[i] % d);
  cur -= mx * mx * (a[i] % d);
  return cur;
}
struct st {
  int i, d;
  bool operator<(const st& o) const { return f(i, d) < f(o.i, o.d); }
};
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) scanf("%lld", a + i);
  priority_queue<st> pq;
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    pq.push({i, 1});
    ans += a[i] * a[i];
  }
  for (int i = n; i < k; i++) {
    auto t = pq.top();
    pq.pop();
    ans -= f(t.i, t.d);
    pq.push({t.i, t.d + 1});
  }
  cout << ans << "\n";
}
