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 cal(int first, int k) {
  long long a = first / k, b = first / k + 1;
  long long val = k - first % k, val2 = first % k;
  return a * a * val + b * b * val2;
}
void solve() {
  int n, k;
  scanf("%d %d", &n, &k);
  long long a[100005];
  int now[1000005];
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
  }
  sort(a, a + n);
  long long ans = 0;
  priority_queue<pair<long long, long long> > pq;
  for (int i = 0; i < n; i++) {
    ans += a[i] * a[i];
    now[i] = 1;
    pq.push(make_pair(cal(a[i], 1) - cal(a[i], 2), i));
  }
  now[n] = 1e9;
  k -= n;
  for (int i = 0; i < k; i++) {
    ans -= pq.top().first;
    int first = pq.top().second;
    pq.pop();
    now[first]++;
    pq.push(make_pair(cal(a[first], now[first]) - cal(a[first], now[first] + 1),
                      first));
  }
  printf("%lld\n", ans);
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
}
