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;
const long long mod = 1000000007;
template <typename T>
istream& operator>>(istream& in, vector<T>& v) {
  for (int i = 0; i < v.size(); i++) {
    in >> v[i];
  }
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> v) {
  for (int i = 0; i < v.size(); i++) {
    out << v[i] << ' ';
  }
  return out;
}
int cnt(long long a) {
  int ans = 0;
  vector<int> b;
  while (a) {
    b.push_back(a % 2);
    a /= 2;
  }
  reverse(b.begin(), b.end());
  for (auto i : b) {
    ans *= 2;
    ans += i;
  }
  return ans;
}
int cnt2(long long a) {
  int res = 0;
  while (a) {
    res += a % 2;
    a /= 2;
  }
  return res;
}
void solve() {
  long long n, p;
  cin >> n >> p;
  for (int i = 1; i < 1000; i++) {
    n -= p;
    if (n < 1) {
      cout << -1 << endl;
      return;
    }
    if (cnt(n) >= i && cnt2(n) <= i) {
      cout << i << endl;
      return;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
