#include <bits/stdc++.h>
using namespace std;
long long M = 1e9 + 7, N = 205001;
long long mm(long long x, long long y) {
  x %= M, y %= M;
  return (x * y) % M;
}
long long po(long long x, long long y) {
  if (!y) return 1;
  long long a = po(x, y / 2) % M;
  if (y % 2) return mm(a, mm(a, x));
  return mm(a, a);
}
long long wt(long long n) {
  long long ans = 0;
  while (n) {
    ans += n % 2;
    n /= 2;
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, p;
  cin >> n >> p;
  for (long long i = 1; 1 < 32 ? i < 32 : i > 32; 1 < 32 ? i += 1 : i -= 1)
    if (n - i * p >= i and wt(n - i * p) <= i) {
      cout << i;
      return 0;
    }
  cout << -1;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long M = 1e9 + 7, N = 205001;
long long mm(long long x, long long y) {
  x %= M, y %= M;
  return (x * y) % M;
}
long long po(long long x, long long y) {
  if (!y) return 1;
  long long a = po(x, y / 2) % M;
  if (y % 2) return mm(a, mm(a, x));
  return mm(a, a);
}
long long wt(long long n) {
  long long ans = 0;
  while (n) {
    ans += n % 2;
    n /= 2;
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, p;
  cin >> n >> p;
  for (long long i = 1; 1 < 32 ? i < 32 : i > 32; 1 < 32 ? i += 1 : i -= 1)
    if (n - i * p >= i and wt(n - i * p) <= i) {
      cout << i;
      return 0;
    }
  cout << -1;
}

```