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 N = 100005;
int n, k, a[N], f[N];
priority_queue<pair<long long, long long> > Q;
long long F(int x, int y) {
  if (x < y) return 1ll << 40;
  long long c = x / y, s1 = x % y, s2 = y - s1;
  return (c + 1) * (c + 1) * s1 + c * c * s2;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &a[i]), f[i] = 1;
  for (int i = (int)(1); i <= (int)(n); i++)
    Q.push(pair<long long, long long>(F(a[i], f[i]) - F(a[i], f[i] + 1), i));
  for (int i = (int)(n + 1); i <= (int)(k); i++) {
    int x = Q.top().second;
    Q.pop();
    ++f[x];
    Q.push(pair<long long, long long>(F(a[x], f[x]) - F(a[x], f[x] + 1), x));
  }
  long long ans = 0;
  for (int i = (int)(1); i <= (int)(n); i++) ans += F(a[i], f[i]);
  cout << ans << endl;
}
