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::chrono;
using namespace std;
const long long mod = 1000000007;
const double pi = 3.14159265359;
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long sqr(long long i) { return i * i; }
long long calc(long long l, long long p) {
  long long w = l / p, p2 = l % p, p1 = p - p2;
  return p1 * (sqr(w)) + p2 * (sqr(w + 1));
}
int32_t main() {
  std::ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  ;
  long long n, k;
  cin >> n >> k;
  long long arr[n];
  long long cost = 0;
  priority_queue<pair<long long, pair<long long, long long>>> pq;
  for (long long i = 0; i < n; i++)
    cin >> arr[i], cost += sqr(arr[i]),
        pq.push({calc(arr[i], 1) - calc(arr[i], 2), {arr[i], 2}});
  for (long long i = 0; i < k - n; i++) {
    cost -= pq.top().first;
    long long p = pq.top().second.second;
    long long l = pq.top().second.first;
    pq.pop();
    pq.push({calc(l, p) - calc(l, p + 1), {l, p + 1}});
  }
  cout << cost << endl;
}
