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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
template <class Iter>
void rprintf(const char *fmt, Iter begin, Iter end) {
  for (bool sp = 0; begin != end; ++begin) {
    if (sp)
      putchar(' ');
    else
      sp = true;
    printf(fmt, *begin);
  }
  putchar('\n');
}
int N, K;
long long A[100011];
int D[100011];
long long calc(long long a, long long d) {
  assert(a >= d);
  long long q = a / d, r = a % d;
  long long ret = 0;
  ret += q * q * (d - r);
  ret += (q + 1) * (q + 1) * r;
  return ret;
}
void MAIN() {
  scanf("%d%d", &N, &K);
  for (int i = 0, i_len = (N); i < i_len; ++i) scanf("%lld", A + i);
  sort(A, A + N);
  priority_queue<pair<long long, int> > Q;
  int cnt = N;
  long long ans = 0;
  for (int i = 0, i_len = (N); i < i_len; ++i) {
    D[i] = 1;
    long long cur = calc(A[i], D[i]);
    ans += cur;
    if (A[i] > D[i]) {
      long long nxt = calc(A[i], D[i] + 1);
      Q.emplace(cur - nxt, i);
    }
  }
  while (cnt < K) {
    long long diff = Q.top().first;
    int idx = Q.top().second;
    Q.pop();
    ans -= diff;
    D[idx]++;
    cnt++;
    if (A[idx] > D[idx]) {
      long long cur = calc(A[idx], D[idx]);
      long long nxt = calc(A[idx], D[idx] + 1);
      Q.emplace(cur - nxt, idx);
    }
  }
  printf("%lld\n", ans);
}
int main() {
  int TC = 1;
  for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
  return 0;
}
