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 int N = 1e5 + 5;
long long f(long long l, long long p) {
  long long fl = l / p;
  return fl * fl * (p - l % p) + (fl + 1) * (fl + 1) * (l % p);
}
int a[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> a[i];
  multiset<array<long long, 3>> pq;
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    ans += f(a[i], 1);
    if (a[i] > 1) pq.insert({f(a[i], 2) - f(a[i], 1), a[i], 1});
  }
  k -= n;
  while (k > 0 && !pq.empty()) {
    auto now = *pq.begin();
    pq.erase(pq.begin());
    ans += now[0];
    if (now[1] > now[2] + 1) {
      pq.insert(
          {f(now[1], now[2] + 2) - f(now[1], now[2] + 1), now[1], now[2] + 1});
    }
    k--;
  }
  cout << ans << '\n';
}
