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 value(int len, int parts) {
  int len1 = len / parts;
  int len2 = len1 + 1;
  int cnt2 = len % parts;
  int cnt1 = parts - cnt2;
  return (long long)cnt1 * len1 * len1 + (long long)cnt2 * len2 * len2;
}
struct S {
  int len;
  int parts;
  long long impr() const { return value(len, parts) - value(len, parts + 1); }
  bool operator<(const S &he) const { return impr() > he.impr(); }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  multiset<S> st;
  int val;
  for (int i = 0; i < n; i++) {
    cin >> val;
    st.insert({val, 1});
  }
  for (int i = 0; i < k - n; i++) {
    S s = *st.begin();
    st.erase(st.begin());
    st.insert(S{s.len, s.parts + 1});
  }
  long long ans = 0;
  for (S s : st) {
    ans += value(s.len, s.parts);
  }
  cout << ans;
}
