#include <bits/stdc++.h>
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
long long int fexp(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000007;
    b /= 2;
    a = a * a % 1000000007;
  }
  return ans;
}
using namespace std;
long long int num_to_bits[16] = {0, 1, 1, 2, 1, 2, 2, 3,
                                 1, 2, 2, 3, 2, 3, 3, 4};
long long int countSetBits(long long int num) {
  long long int nibble = 0;
  if (0 == num) return num_to_bits[0];
  nibble = num & 0xf;
  return num_to_bits[nibble] + countSetBits(num >> 4);
}
int main() {
  long long int t;
  t = 1;
  while (t--) {
    long long int n, p;
    cin >> n >> p;
    if (n <= p) {
      cout << "-1";
      break;
    }
    long long int k = 1;
    long long int m = n - p;
    while (k < countSetBits(m) || m < k) {
      m -= p;
      k++;
      if (m < 0) {
        k = -1;
        break;
      }
    }
    cout << k << endl;
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
long long int fexp(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000007;
    b /= 2;
    a = a * a % 1000000007;
  }
  return ans;
}
using namespace std;
long long int num_to_bits[16] = {0, 1, 1, 2, 1, 2, 2, 3,
                                 1, 2, 2, 3, 2, 3, 3, 4};
long long int countSetBits(long long int num) {
  long long int nibble = 0;
  if (0 == num) return num_to_bits[0];
  nibble = num & 0xf;
  return num_to_bits[nibble] + countSetBits(num >> 4);
}
int main() {
  long long int t;
  t = 1;
  while (t--) {
    long long int n, p;
    cin >> n >> p;
    if (n <= p) {
      cout << "-1";
      break;
    }
    long long int k = 1;
    long long int m = n - p;
    while (k < countSetBits(m) || m < k) {
      m -= p;
      k++;
      if (m < 0) {
        k = -1;
        break;
      }
    }
    cout << k << endl;
  }
}

```