#include <bits/stdc++.h>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")
using namespace std;
mt19937_64 rnd(time(0));
long long gcd(long long a, long long b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
const long long N = 2e5 + 10, INF = 1e9 + 10, K = 1e6 + 10;
long long a[N];
long long cnt[K];
long long popcnt(long long x) {
  long long res = 0;
  while (x) {
    if (x ^ (x & (x - 1))) {
      ++res;
    }
    x &= (x - 1);
  }
  return res;
}
void solve() {
  long long n, p;
  cin >> n >> p;
  for (long long i = 1; i < 63; ++i) {
    long long x = n - i * p;
    if (x <= 0) {
      continue;
    }
    long long cnt = popcnt(x);
    if (i >= cnt && i <= x) {
      cout << i;
      exit(0);
    }
  }
  cout << -1;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")
using namespace std;
mt19937_64 rnd(time(0));
long long gcd(long long a, long long b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
const long long N = 2e5 + 10, INF = 1e9 + 10, K = 1e6 + 10;
long long a[N];
long long cnt[K];
long long popcnt(long long x) {
  long long res = 0;
  while (x) {
    if (x ^ (x & (x - 1))) {
      ++res;
    }
    x &= (x - 1);
  }
  return res;
}
void solve() {
  long long n, p;
  cin >> n >> p;
  for (long long i = 1; i < 63; ++i) {
    long long x = n - i * p;
    if (x <= 0) {
      continue;
    }
    long long cnt = popcnt(x);
    if (i >= cnt && i <= x) {
      cout << i;
      exit(0);
    }
  }
  cout << -1;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```