#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-7;
const string alpha = "abcdefghijklmnopqrstuvwxyz";
int dx[4] = {1, 0, 0, -1};
int dy[4] = {0, -1, 1, 0};
long long power(long long x, long long y) {
  long long ans = 1;
  x = x;
  while (y > 0) {
    if (y & 1) ans = (ans * x);
    y = y >> 1;
    x = (x * x);
  }
  return ans;
}
long long sqr(long long x) { return x * x; }
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <class T>
inline void chmax(T &x, T y) {
  x = max((x), (y));
}
template <class T>
inline void chmin(T &x, T y) {
  x = min((x), (y));
}
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T powm(T a, T b, long long m) {
  T cnt = 1;
  while (b > 0) {
    if (b % 2 == 1) cnt = (cnt * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return cnt % m;
}
long long ncr(long long n, long long r) {
  long long ans = 1;
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  for (long long i = 0; i < r; i++) {
    ans *= (n - i);
    ans /= (i + 1);
  }
  return ans;
}
const int MOD = 1000000007;
const int N = 1005;
struct compare {
  bool operator()(const int &a, const int &b) const { return a > b; }
};
void prec() {}
void run() {
  long long n, p;
  cin >> n >> p;
  for (int i = 1; i <= 40; ++i) {
    long long now = n - (p * i), cnt = 0;
    for (int j = 0; j <= 45; ++j) {
      if ((now >> j) & 1) cnt++;
    }
    if (cnt <= i and now >= 0 and i <= now) {
      cout << i;
      return;
    }
  }
  cout << -1;
}
int32_t main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int tc = 1;
  prec();
  for (long long i = 0; i < tc; ++i) run();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const long double eps = 1e-7;
const string alpha = "abcdefghijklmnopqrstuvwxyz";
int dx[4] = {1, 0, 0, -1};
int dy[4] = {0, -1, 1, 0};
long long power(long long x, long long y) {
  long long ans = 1;
  x = x;
  while (y > 0) {
    if (y & 1) ans = (ans * x);
    y = y >> 1;
    x = (x * x);
  }
  return ans;
}
long long sqr(long long x) { return x * x; }
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <class T>
inline void chmax(T &x, T y) {
  x = max((x), (y));
}
template <class T>
inline void chmin(T &x, T y) {
  x = min((x), (y));
}
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T powm(T a, T b, long long m) {
  T cnt = 1;
  while (b > 0) {
    if (b % 2 == 1) cnt = (cnt * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return cnt % m;
}
long long ncr(long long n, long long r) {
  long long ans = 1;
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  for (long long i = 0; i < r; i++) {
    ans *= (n - i);
    ans /= (i + 1);
  }
  return ans;
}
const int MOD = 1000000007;
const int N = 1005;
struct compare {
  bool operator()(const int &a, const int &b) const { return a > b; }
};
void prec() {}
void run() {
  long long n, p;
  cin >> n >> p;
  for (int i = 1; i <= 40; ++i) {
    long long now = n - (p * i), cnt = 0;
    for (int j = 0; j <= 45; ++j) {
      if ((now >> j) & 1) cnt++;
    }
    if (cnt <= i and now >= 0 and i <= now) {
      cout << i;
      return;
    }
  }
  cout << -1;
}
int32_t main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int tc = 1;
  prec();
  for (long long i = 0; i < tc; ++i) run();
  return 0;
}

```