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 sq(long long x) { return x * x; }
long long first(long long l, long long p) {
  long long w = l / p;
  long long p2 = l % p;
  long long p1 = p - p2;
  return p1 * sq(w) + p2 * sq(w + 1);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long int> a(n);
  for (auto &it : a) cin >> it;
  priority_queue<pair<long long, pair<long long int, long long int>>> second;
  long long ans = 0, d = k - n;
  for (auto it : a) {
    second.push({first(it, 1) - first(it, 2), {it, 2}});
    ans += sq(it);
  }
  while (d) {
    pair<long long, pair<long long int, long long int>> v = second.top();
    second.pop();
    ans -= v.first;
    second.push({first(v.second.first, v.second.second) -
                     first(v.second.first, v.second.second + 1),
                 {v.second.first, v.second.second + 1}});
    d--;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(20);
  long long t = 1;
  while (t--) {
    solve();
    cout << endl;
  }
  cin.get();
  return 0;
}
