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 int f(long long int l, long long int p) {
  long long int g = l % p;
  return g * ((l / p) + 1) * ((l / p) + 1) + (p - g) * (l / p) * (l / p);
}
int main() {
  long long int n, k, e = 1000000;
  cin >> n >> k;
  long long int a[n], b[n];
  set<long long int> s;
  for (long long int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    b[i] = 1;
    if (a[i] >= b[i] + 1) {
      long long int d = f(a[i], 1) - f(a[i], 2);
      d *= e;
      d += i;
      s.insert(d);
    }
  }
  for (long long int i = 0; i < k - n; i++) {
    set<long long int>::iterator it;
    it = s.end();
    it--;
    long long int g = (*it) % e;
    b[g]++;
    s.erase(it);
    if (a[g] >= b[g] + 1) {
      long long int d = f(a[g], b[g]) - f(a[g], b[g] + 1);
      d *= e;
      d += g;
      s.insert(d);
    }
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    ans += f(a[i], b[i]);
  }
  cout << ans << endl;
}
