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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long Mod = 1e9 + 7;
const long long inf = 1e18;
long long sq(long long x) { return (long long)x * (long long)x; }
long long cal(long long N, long long K) {
  long long p = N / K;
  long long rem = N % K;
  long long h = rem;
  long long l = K - rem;
  return (h * sq(p + 1)) + (l * sq(p));
};
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  long long ans = 0;
  priority_queue<tuple<long long, long long, long long>> pq;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    ans += cal(x, 1);
    pq.push({cal(x, 1) - cal(x, 2), x, 2});
  }
  for (long long i = 0; i < k - n; i++) {
    auto [c, a, b] = pq.top();
    pq.pop();
    ans -= c;
    pq.push({cal(a, b) - cal(a, b + 1), a, b + 1});
  }
  cout << ans << '\n';
}
