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 n, k, v[100000 + 10];
priority_queue<pair<long long, pair<long long, long long> > > pq;
long long calcBuc(int idx, long long nrbuc) {
  if (nrbuc > v[idx]) return 2000000000;
  long long val1 = v[idx] / nrbuc, val2 = val1 + 1;
  long long cnt2 = v[idx] % nrbuc, cnt1 = nrbuc - cnt2;
  return val1 * val1 * cnt1 + val2 * val2 * cnt2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    pq.push({calcBuc(i, 1) - calcBuc(i, 2), {i, 1}});
  }
  k -= n;
  while (k--) {
    pair<long long, pair<long long, long long> > x = pq.top();
    pq.pop();
    pq.push({calcBuc(x.second.first, x.second.second + 1) -
                 calcBuc(x.second.first, x.second.second + 2),
             {x.second.first, x.second.second + 1}});
  }
  long long ans = 0;
  while (!pq.empty()) {
    pair<long long, pair<long long, long long> > x = pq.top();
    pq.pop();
    ans = ans + calcBuc(x.second.first, x.second.second);
  }
  cout << ans << '\n';
  return 0;
}
