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 <typename X>
ostream& operator<<(ostream& x, const vector<X>& v) {
  for (long long i = 0; i < v.size(); ++i) x << v[i] << " ";
  return x;
}
template <typename X>
ostream& operator<<(ostream& x, const set<X>& v) {
  for (auto it : v) x << it << " ";
  return x;
}
template <typename X, typename Y>
ostream& operator<<(ostream& x, const pair<X, Y>& v) {
  x << v.first << " " << v.second;
  return x;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const map<T, S>& v) {
  for (auto it : v) os << it.first << "=>" << it.second << endl;
  return os;
}
const long long N = 1e5 + 10;
long long dp[N];
long long n, k;
long long a[N];
long long f(long long a, long long b) {
  long long t = a / b;
  long long r = a % b;
  long long val = t * t * (b - r) + (t + 1) * (t + 1) * r;
  return val;
}
class ComparisonClass {
 public:
  bool operator()(pair<long long, long long> a, pair<long long, long long> b) {
    return f(a.first, a.second) - f(a.first, a.second + 1) <
           f(b.first, b.second) - f(b.first, b.second + 1);
  }
};
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  priority_queue<pair<long long, long long>,
                 vector<pair<long long, long long> >, ComparisonClass>
      q;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    q.push({a[i], 1});
  }
  long long d = k - n;
  while (q.size() > 0 and d--) {
    auto x = q.top();
    q.pop();
    42;
    q.push({x.first, x.second + 1});
  }
  long long ans = 0;
  while (q.size() > 0) {
    auto x = q.top();
    q.pop();
    42;
    ans += f(x.first, x.second);
  }
  cout << ans << endl;
  return 0;
}
