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, a[100005];
long long ans;
struct QWQ {
  int num, cnt;
  long long val, sum;
  bool friend operator<(QWQ x, QWQ y) { return x.val < y.val; }
} temp;
priority_queue<QWQ> q;
inline long long cal(int x, int y, int z) {
  long long ans1 = 1LL * x * (x - 1) / 2 * (z + 1) * (z + 1);
  long long ans2 = 1LL * y * (y - 1) / 2 * z * z;
  long long ans3 = 1LL * x * y * z * (z + 1);
  return ans1 + ans2 + ans3;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    ans += 1LL * a[i] * a[i];
    temp.num = a[i], temp.cnt = 2, temp.sum = 0;
    int x = a[i] % 2, y = 2 - a[i] % 2, z = a[i] / 2;
    temp.val = cal(x, y, z);
    q.push(temp);
  }
  for (int i = n + 1; i <= k; ++i) {
    temp = q.top();
    q.pop();
    long long val = temp.val, sum = temp.sum + val;
    int cnt = temp.cnt + 1, num = temp.num;
    ans -= val * 2;
    int x = num % cnt, y = cnt - num % cnt, z = num / cnt;
    temp.cnt = cnt, temp.sum = sum, temp.val = cal(x, y, z) - sum;
    q.push(temp);
  }
  printf("%lld\n", ans);
  return 0;
}
