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 ≤ n ≤ 10^9, -1000 ≤ p ≤ 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.
#include <bits/stdc++.h>
using namespace std;
template <class R>
inline void Readin(R &K) {
  bool F = K = 0;
  char C = getchar();
  while (C < '0' || C > '9') {
    if (C == '-') F = 1;
    C = getchar();
  }
  while (C <= '9' && C >= '0') K = (K << 1) + (K << 3) + C - '0', C = getchar();
  if (F) K = -K;
}
template <class R>
inline bool Checkmax(R &A, R B) {
  return A < B ? A = B, 1 : 0;
}
template <class R>
inline bool Checkmin(R &A, R B) {
  return A > B ? A = B, 1 : 0;
}
inline int Bit(int A) {
  if (A <= 0) return 100000;
  int Ans = 0;
  while (A) {
    Ans += A & 1;
    A >>= 1;
  }
  return Ans;
}
int main() {
  int A, B;
  Readin(A);
  Readin(B);
  for (register int i = 1; i <= 32; ++i)
    if (Bit(A - i * B) <= i and A - i * B >= i) return not printf("%d\n", i);
  return not printf("-1\n");
}
