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;
using namespace std;
const int maxn = 100000 + 10;
const long long inf = 1e18;
int n, k;
long long a[maxn];
inline long long calc(long long l, long long p) {
  if (p <= 0) return inf;
  long long tmp = (l / p) * (l / p) * (p - l % p);
  tmp += ((l + p - 1) / p) * ((l + p - 1) / p) * (l % p);
  return tmp;
}
struct node {
  long long len, cnt;
  long long getval() const { return calc(len, cnt) - calc(len, cnt + 1); }
  bool operator<(const node& o) const { return getval() < o.getval(); }
};
inline long long solve() {
  priority_queue<node> q;
  long long ans = 0;
  for (auto i = (1); i <= (n); ++i) q.push({a[i], 1}), ans += a[i] * a[i];
  k -= n;
  for (; k;) {
    auto e = q.top();
    q.pop();
    ans -= calc(e.len, e.cnt);
    ans += calc(e.len, ++e.cnt);
    q.push(e);
    k--;
  }
  return ans;
}
int main(void) {
  scanf("%d%d", &n, &k);
  for (auto i = (1); i <= (n); ++i) scanf("%lld", &(a[i]));
  sort(a + 1, a + 1 + n);
  printf("%lld\n", solve());
  return 0;
}
