#include <bits/stdc++.h>
using namespace std;
const long long int inf = 1e18;
const long long int mod = 1e9 + 7;
const long long int MOD = 998244353;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug &operator<<(const c &) {
    return *this;
  }
};
vector<char *> tokenizer(const char *args) {
  char *token = new char[111];
  strcpy(token, args);
  token = strtok(token, ", ");
  vector<char *> v({token});
  while ((token = strtok(NULL, ", "))) v.push_back(token);
  return reverse(v.begin(), v.end()), v;
}
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wunused-parameter"
void debugg(vector<char *> args) { cerr << "\b\b "; }
#pragma clang diagnostic pop
template <typename Head, typename... Tail>
void debugg(vector<char *> args, Head H, Tail... T) {
  debug() << " [" << args.back() << ": " << H << "] ";
  args.pop_back();
  debugg(args, T...);
}
template <typename T>
T power(T a, T b) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    T res = (power(a, b / 2));
    if (b % 2) {
      return (res * res * a);
    } else {
      return res * res;
    }
  }
}
template <typename T>
T power(T a, T b, T modulo) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    T res = (power(a, b / 2, modulo) % modulo);
    if (b % 2) {
      return ((((res % modulo) * (res % modulo)) % modulo) * (a % modulo)) %
             modulo;
    } else {
      return ((res % modulo) * (res % modulo)) % modulo;
    }
  }
}
template <typename T>
T gcd(T a, T b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
void factorial(vector<long long int> &fact, long long n) {
  fact.resize(n + 1, 1);
  fact[0] = 1;
  fact[1] = 1;
  for (int i = 2; i <= n; ++i) {
    fact[i] = ((fact[i - 1] % mod) * (i % mod)) % mod;
  }
}
long long mod_inv(long long a) {
  return (power<long long>(a, mod - 2, mod)) % mod;
}
long long ncr(long long n, long long r, vector<long long int> &fact) {
  if (r > n or n < 0 or r < 0) return 0LL;
  return (((fact[n] % mod) * (mod_inv(fact[n - r]) % mod)) % mod *
          (mod_inv(fact[r]) % mod)) %
         mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n, p;
  cin >> n >> p;
  long long ans = inf;
  for (long long i = 0; i <= 32; ++i) {
    long long s = n - i * p;
    if (s < i) continue;
    long long d = __builtin_popcountll(s);
    if (i >= d) {
      {};
      ans = min(ans, i);
    }
  }
  if (ans == inf) ans = -1;
  cout << ((ans)) << "\n";
  ;
  return 0;
}

### 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;
const long long int inf = 1e18;
const long long int mod = 1e9 + 7;
const long long int MOD = 998244353;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug &operator<<(const c &) {
    return *this;
  }
};
vector<char *> tokenizer(const char *args) {
  char *token = new char[111];
  strcpy(token, args);
  token = strtok(token, ", ");
  vector<char *> v({token});
  while ((token = strtok(NULL, ", "))) v.push_back(token);
  return reverse(v.begin(), v.end()), v;
}
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wunused-parameter"
void debugg(vector<char *> args) { cerr << "\b\b "; }
#pragma clang diagnostic pop
template <typename Head, typename... Tail>
void debugg(vector<char *> args, Head H, Tail... T) {
  debug() << " [" << args.back() << ": " << H << "] ";
  args.pop_back();
  debugg(args, T...);
}
template <typename T>
T power(T a, T b) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    T res = (power(a, b / 2));
    if (b % 2) {
      return (res * res * a);
    } else {
      return res * res;
    }
  }
}
template <typename T>
T power(T a, T b, T modulo) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    T res = (power(a, b / 2, modulo) % modulo);
    if (b % 2) {
      return ((((res % modulo) * (res % modulo)) % modulo) * (a % modulo)) %
             modulo;
    } else {
      return ((res % modulo) * (res % modulo)) % modulo;
    }
  }
}
template <typename T>
T gcd(T a, T b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
void factorial(vector<long long int> &fact, long long n) {
  fact.resize(n + 1, 1);
  fact[0] = 1;
  fact[1] = 1;
  for (int i = 2; i <= n; ++i) {
    fact[i] = ((fact[i - 1] % mod) * (i % mod)) % mod;
  }
}
long long mod_inv(long long a) {
  return (power<long long>(a, mod - 2, mod)) % mod;
}
long long ncr(long long n, long long r, vector<long long int> &fact) {
  if (r > n or n < 0 or r < 0) return 0LL;
  return (((fact[n] % mod) * (mod_inv(fact[n - r]) % mod)) % mod *
          (mod_inv(fact[r]) % mod)) %
         mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n, p;
  cin >> n >> p;
  long long ans = inf;
  for (long long i = 0; i <= 32; ++i) {
    long long s = n - i * p;
    if (s < i) continue;
    long long d = __builtin_popcountll(s);
    if (i >= d) {
      {};
      ans = min(ans, i);
    }
  }
  if (ans == inf) ans = -1;
  cout << ((ans)) << "\n";
  ;
  return 0;
}

```