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 = 1e5 + 5;
long long n, m;
long long shu[maxn];
long long ans;
struct nod {
  long long x, k, id;
  nod() {}
  nod(long long a1, long long a2, long long a3) { x = a1, k = a2, id = a3; }
  friend bool operator<(nod shu1, nod shu2) { return shu1.x < shu2.x; }
};
priority_queue<nod> dui;
inline long long f(long long x, long long k) {
  return (x / k + 1) * (x / k + 1) * (x % k) + (x / k) * (x / k) * (k - x % k);
}
signed main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  m -= n;
  for (long long i = 1; i <= n; ++i) {
    cin >> shu[i];
    ans += shu[i] * shu[i];
    dui.push(nod(f(shu[i], 1) - f(shu[i], 2), 1, i));
  }
  while (m--) {
    nod now = dui.top();
    dui.pop();
    long long id = now.id, k = now.k + 1;
    ans -= now.x;
    dui.push(nod(f(shu[id], k) - f(shu[id], k + 1), k, id));
  }
  cout << ans;
}
