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 N = 1e5 + 1;
long long n, k, ans = 0, a[N], part[N];
priority_queue<pair<long long, long long>> pq;
long long divide(long long val, long long k) {
  if (val <= k) return val;
  long long x = val / k, y = val % k, z = k - y;
  return z * x * x + y * (x + 1) * (x + 1);
}
signed main() {
  cin >> n >> k;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    part[i] = 1;
    ans += a[i] * a[i];
    pq.push({divide(a[i], part[i]) - divide(a[i], part[i] + 1), i});
  }
  for (long long i = 0; i < k - n; ++i) {
    auto pick = pq.top();
    pq.pop();
    ans -= pick.first;
    ++part[pick.second];
    pq.push({divide(a[pick.second], part[pick.second]) -
                 divide(a[pick.second], part[pick.second] + 1),
             pick.second});
  }
  cout << ans;
}
