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 power(long long x, long long y, long long m) {
  long long r = 1;
  for (; y; y >>= 1) {
    if (y & 1) r = r * x % m;
    x = x * x % m;
  }
  return r;
}
const long long N = 55, M = 1e18 + 7;
long long dv(long long x, long long p) {
  long long c = x % p, y = x / p;
  return c * (y + 1) * (y + 1) + (p - c) * y * y;
}
int run_test() {
  long long n, k;
  cin >> n >> k;
  long long a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  priority_queue<pair<long long, pair<long long, long long> > > q;
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    ans += a[i] * a[i];
    q.push({a[i] * a[i] - dv(a[i], 2LL), {a[i], 2LL}});
  }
  while (n < k) {
    pair<long long, pair<long long, long long> > p = q.top();
    q.pop();
    ans -= p.first;
    q.push({dv(p.second.first, p.second.second) -
                dv(p.second.first, p.second.second + 1),
            {p.second.first, p.second.second + 1}});
    n++;
  }
  cout << ans;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long t;
  t = 1;
  while (t--) {
    run_test();
  }
}
