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 maxn = 1e5 + 5;
struct node {
  int val;
  int cnt;
  int i;
  inline long long cal(const long long& x, const long long& cnt) const {
    long long num = x / cnt;
    long long res = x % cnt;
    return 1ll * (cnt - res) * num * num + 1ll * res * (num + 1) * (num + 1);
  }
  inline bool operator<(const node& bb) const {
    return cal(val, cnt) - cal(val, cnt + 1) <
           cal(bb.val, bb.cnt) - cal(bb.val, bb.cnt + 1);
  }
};
int a[maxn];
int cnt[maxn];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  priority_queue<node> q;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    cnt[i] = 1;
    q.push({a[i], 1, i});
  }
  int n_ = n;
  while (n_ < k) {
    node tp = q.top();
    q.pop();
    cnt[tp.i]++;
    q.push({tp.val, tp.cnt + 1, tp.i});
    ++n_;
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    long long num = a[i] / cnt[i];
    long long res = a[i] % cnt[i];
    ans += 1ll * (cnt[i] - res) * num * num + 1ll * res * (num + 1) * (num + 1);
  }
  printf("%lld\n", ans);
  return 0;
}
