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 = 233333;
long long n, k, sum;
long long a[maxn];
struct DD {
  DD() {}
  DD(long long _d, long long _l, long long _w) { d = _d, l = _l, w = _w; }
  long long d, l, w;
  bool operator<(const DD& b) const { return w < b.w; }
  void output() { printf("len: %lld d: %lld w: %lld\n", l, d, w); }
};
inline long long sqr(long long x) { return x * x; }
inline long long calc(long long l, long long d) {
  long long r = l % d, q = l / d;
  return r * sqr(q + 1) + (d - r) * sqr(q);
}
bool chk(long long mid) {
  long long m = sum - mid;
  if (m <= 0) return true;
  priority_queue<DD> q;
  for (int i = 1; i <= n; ++i) {
    q.push(DD(1, a[i], calc(a[i], 1) - calc(a[i], 2)));
  }
  long long sz = k - n;
  DD cur;
  for (long long i = 1; i <= sz; ++i) {
    cur = q.top();
    q.pop();
    m -= cur.w;
    ++cur.d;
    cur.w = calc(cur.l, cur.d) - calc(cur.l, cur.d + 1);
    if (m <= 0) return true;
    q.push(cur);
  }
  return false;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
  }
  sum = 0;
  for (int i = 1; i <= n; ++i) {
    sum += a[i] * a[i];
  }
  long long le = 0, ri = 1e18;
  long long mid = (le + ri) >> 1;
  while (le < ri) {
    if (chk(mid)) {
      ri = mid;
    } else {
      le = mid + 1;
    }
    mid = (le + ri) >> 1;
  }
  printf("%lld\n", mid);
}
