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;
int N, K;
long long arr[100005];
priority_queue<pair<long long, int>> pq;
int cnt[100005];
long long calc(long long n, long long k) {
  long long r = n % k;
  return (n / k) * (n / k) * ((k - r)) + (n / k + 1) * (n / k + 1) * r;
}
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 >> arr[i];
    cnt[i] = 1;
    ans += 1LL * arr[i] * arr[i];
    if (arr[i] - 1) {
      long long val = calc(arr[i], 1) - calc(arr[i], 2);
      pq.push(make_pair(val, i));
    }
  }
  while (pq.size() && N < K) {
    auto p = pq.top();
    ans -= p.first;
    pq.pop();
    N++;
    if (cnt[p.second] == arr[p.second]) {
      continue;
    }
    cnt[p.second]++;
    long long val = calc(arr[p.second], cnt[p.second]) -
                    calc(arr[p.second], cnt[p.second] + 1);
    pq.push(make_pair(val, p.second));
  }
  cout << ans;
}
