P-binary

Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer $p$ (which may be positive, negative, or zero). To combine their tastes, they invented $p$-binary numbers of the form $2^x + p$, where $x$ is a non-negative integer.

For example, some $-9$-binary (“minus nine” binary) numbers are: $-8$ (minus eight), $7$ and $1015$ ($-8=2^0-9$, $7=2^4-9$, $1015=2^{10}-9$).

The boys now use $p$-binary numbers to represent everything. They now face a problem: given a positive integer $n$, what’s the smallest number of $p$-binary numbers (not necessarily distinct) they need to represent $n$ as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.

For example, if $p=0$ we can represent $7$ as $2^0 + 2^1 + 2^2$.

And if $p=-9$ we can represent $7$ as one number $(2^4-9)$.

Note that negative $p$-binary numbers are allowed to be in the sum (see the Notes section for an example).Input

The only line contains two integers $n$ and $p$ ($1 \leq n \leq 10^9$, $-1000 \leq p \leq 1000$).Output

If it is impossible to represent $n$ as the sum of any number of $p$-binary numbers, print a single integer $-1$. Otherwise, print the smallest possible number of summands.

Examples input

24 0

output

2

input

24 1

output

3

input

24 -1

output

4

input

4 -7

output

2

input

1 1

output

-1

Note

$0$-binary numbers are just regular binary powers, thus in the first sample case we can represent $24 = (2^4 + 0) + (2^3 + 0)$.

In the second sample case, we can represent $24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1)$.

In the third sample case, we can represent $24 = (2^4 – 1) + (2^2 – 1) + (2^2 – 1) + (2^2 – 1)$. Note that repeated summands are allowed.

In the fourth sample case, we can represent $4 = (2^4 – 7) + (2^1 – 7)$. Note that the second summand is negative, which is allowed.

In the fifth sample case, no representation is possible.

Solution:

#include <bits/stdc++.h>

using namespace std;

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, p;
cin >> n >> p;
const int inf = (int) 1e9;
int ans = inf;
for (int c = 1; c <= 100000; c++) {
    int num = n - p * c;
    if (num >= c) {
int pc = __builtin_popcount(num);
if (pc <= c) {
        ans = c;
        break;
      }
    }
  }
  cout << (ans == inf ? -1 : ans) << '\n';
  return 0;
}