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 rr[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long cc[] = {1, 1, 0, -1, -1, -1, 0, 1};
const long long inf = 1e18;
const long long mod = 1e9 + 7;
const long long N = 300050;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long test = 1;
  while (test--) {
    long long n, k, K;
    cin >> n >> k;
    k -= n;
    K = k;
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a, a + n, greater<long long>());
    priority_queue<tuple<long long, long long, long long> > q;
    auto find = [&](long long l, long long p) {
      long long tot =
          (l % p) * (l / p + 1) * (l / p + 1) + (p - l % p) * (l / p) * (l / p);
      return tot;
    };
    for (long long i = 0; i < min(k, n); i++) {
      long long diff = find(a[i], 1LL) - find(a[i], 2LL);
      q.push({diff, a[i], 1});
    }
    long long ans = 0;
    for (long long i = min(k, n); i < n; i++) {
      ans += a[i] * a[i];
    }
    while (k--) {
      auto x = q.top();
      q.pop();
      long long D, A, P;
      tie(D, A, P) = x;
      D = find(A, P + 1) - find(A, P + 2);
      q.push({D, A, P + 1});
    }
    while (!q.empty()) {
      long long D, A, P;
      tie(D, A, P) = q.top();
      q.pop();
      ans += find(A, P);
    }
    cout << ans << '\n';
  }
  return 0;
}
