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;
const long long N = 2e5;
long long n, k;
long long a[N];
inline long long f(long long x, long long k) {
  if (x < k) return 100000;
  long long w = x / k;
  long long las = x - w * k;
  return w * w * (k - las) + (w + 1) * (w + 1) * las;
}
priority_queue<pair<long long, long long> > q;
long long cuts[N];
long long ans = 0;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n; i++) {
    q.push(make_pair(a[i] * a[i] - f(a[i], 2), i));
    cuts[i] = 2;
    ans += a[i] * a[i];
  }
  for (long long t = 1; t <= k - n; t++) {
    pair<long long, long long> tp = q.top();
    long long pos = tp.second;
    long long w = tp.first;
    ans -= w;
    q.pop();
    q.push(make_pair(f(a[pos], cuts[pos]) - f(a[pos], cuts[pos] + 1), pos));
    cuts[pos]++;
  }
  cout << ans << endl;
}
