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 long long M = 1e5 + 10;
long long a[M], cnt[M], n, k, ans;
long long f(long long x, long long y) {
  return (y - (x % y)) * (x / y) * (x / y) +
         (x % y) * (x / y + 1) * (x / y + 1);
}
struct node {
  long long pos, val;
};
bool operator<(const node &a, const node &b) { return a.val > b.val; }
priority_queue<node> q;
inline long long read() {
  char c = getchar();
  long long x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
signed main() {
  n = read(), k = read();
  for (long long i = 1; i <= n; i++)
    a[i] = read(), cnt[i] = 1, q.push((node){i, f(a[i], 2) - a[i] * a[i]}),
    ans += a[i] * a[i];
  for (long long i = 1; i <= k - n; i++) {
    node p = q.top();
    q.pop();
    ans += p.val, cnt[p.pos]++;
    q.push(
        (node){p.pos, f(a[p.pos], cnt[p.pos] + 1) - f(a[p.pos], cnt[p.pos])});
  }
  cout << ans << endl;
  return 0;
}
