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 int mod = 998244353;
long long t, n, k, a[100005], d[100005];
long long cat(long long k, long long p) {
  long long n = a[p];
  long long res = n / k;
  long long r = n % k;
  return res * res * (k - r) + (res + 1) * (res + 1) * r;
}
priority_queue<pair<long long, long long> > q;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    d[i] = 1;
    q.push(pair<long long, long long>(cat(1, i) - cat(2, i), i));
  }
  k -= n;
  for (int i = 1; i <= k; i++) {
    pair<long long, long long> u = q.top();
    q.pop();
    d[u.second]++;
    q.push(pair<long long, long long>(
        cat(d[u.second], u.second) - cat(d[u.second] + 1, u.second), u.second));
  }
  while (!q.empty()) {
    pair<long long, long long> u = q.top();
    ans += cat(d[u.second], u.second);
    q.pop();
  }
  cout << ans;
}
