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 + 5;
const long long INF = 1e9 + 5;
const long long mod = 1e9 + 7;
long long n, k, ans;
priority_queue<pair<long long, pair<long long, long long> > > pq;
long long calc(long long x, long long y) {
  long long d = x / y;
  long long r = x % y;
  return (y - r) * d * d + r * (d + 1) * (d + 1);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    ans += x * x;
    pq.push({calc(x, 1) - calc(x, 2), {x, 2}});
  }
  for (long long i = n; i < k; i++) {
    long long val = pq.top().first;
    long long x = pq.top().second.first;
    long long y = pq.top().second.second;
    pq.pop();
    ans -= val;
    pq.push({calc(x, y) - calc(x, y + 1), {x, y + 1}});
  }
  cout << ans << "\n";
}
