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>
#pragma GCC omptimize("unroll-loops")
#pragma GCC omptimize("ПивоНаПолу))))")
#pragma GCC optimize("no-stack-protector")
#pragma comment(linker, "/STACK:1000000000")
using namespace std;
long long fpow(long long x, long long y) {
  if (!y) return 1;
  if (y == 1) return x;
  if (y % 2)
    return fpow(x, y - 1) * (x);
  else {
    long long d = fpow(x, y / 2);
    return d * d;
  }
}
const int mod = 1e9 + 7;
const int INF32 = 1009000999;
const long long INF64 = 4e18;
const int sz = 6e6;
const long double eps = 1e-12;
long long n, p, d;
long long h[sz];
long long pws[sz];
bool check(long long s, long long r) {
  int c = __builtin_popcount(r);
  return r >= s && s >= c && r > 0;
}
void Solve() {
  cin >> n >> p;
  for (int i = 0; i < sz; i++) {
    if (check((i + 1), (n - ((i + 1) * p)))) {
      cout << (i + 1);
      return;
    }
  }
  cout << -1;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int q = 1;
  while (q--) Solve();
}
