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;
using i64 = int64_t;
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  vector<i64> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  auto calcScore = [](i64 x, i64 k) -> i64 {
    if (k > x) return 1e18;
    i64 a = x / k;
    i64 ce = x % k;
    return a * a * (k - ce) + (a + 1) * (a + 1) * ce;
  };
  vector<i64> c(n, 1);
  vector<i64> score(n);
  i64 ret = 0;
  set<pair<i64, int>> S;
  for (int i = 0; i < n; ++i) {
    ret += calcScore(a[i], c[i]);
    score[i] = calcScore(a[i], c[i] + 1) - calcScore(a[i], c[i]);
    S.emplace(score[i], i);
  }
  for (int it = 0; it < k - n; ++it) {
    int i = S.begin()->second;
    S.erase(S.begin());
    ret += score[i];
    c[i]++;
    score[i] = calcScore(a[i], c[i] + 1) - calcScore(a[i], c[i]);
    S.emplace(score[i], i);
  }
  cout << ret << '\n';
  return 0;
}
