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 go(long long num, long long parts) {
  long long len1 = num / parts;
  long long len2 = len1 + 1;
  long long rem = num % parts;
  parts -= rem;
  return (parts * len1 * len1) + (rem * len2 * len2);
}
struct S {
  long long len, parts;
  long long imporvement() const { return go(len, parts) - go(len, parts + 1); }
  bool operator<(const S& s) const { return imporvement() > s.imporvement(); }
};
int32_t main() {
  long long n, k;
  cin >> n >> k;
  multiset<S> q;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    q.insert({x, 1});
  }
  long long part = n;
  while (part < k) {
    S s = *q.begin();
    q.erase(q.begin());
    q.insert({s.len, s.parts + 1});
    part++;
  }
  long long ans = 0;
  for (auto each : q) {
    ans += go(each.len, each.parts);
  }
  cout << ans << endl;
  return 0;
}
