#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 2;
const int MAXM = 1e5 + 10;
long long inline in() {
  long long x = 0, flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * flag;
}
int head[MAXN], nume;
struct Adj {
  int nex, to, w;
} adj[MAXM << 1];
void addedge(int from, int to, int w) {
  adj[++nume] = (Adj){head[from], to, w};
  head[from] = nume;
}
void link(int u, int v, int w) {
  addedge(u, v, w);
  addedge(v, u, w);
}
long long n, m;
long long p[60];
int main() {
  n = in();
  m = in();
  p[0] = 1;
  for (int i = 1; i <= 50; ++i) {
    p[i] = p[i - 1] * 2;
  }
  long long ans = 1e18;
  for (long long i = 1; i <= 1000000; ++i) {
    long long now = n - i * m;
    long long need = 0, needm = 0;
    for (int j = 50; j >= 0; --j) {
      if (p[j] > now) continue;
      ++need;
      now -= p[j];
      needm += p[j];
    }
    if (now != 0) continue;
    if (need <= i && i <= needm) {
      ans = min(ans, i);
      break;
    }
  }
  if (ans == 1000000000000000000)
    printf("-1");
  else
    printf("%lld\n", ans);
  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 int MAXN = 1e6 + 2;
const int MAXM = 1e5 + 10;
long long inline in() {
  long long x = 0, flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * flag;
}
int head[MAXN], nume;
struct Adj {
  int nex, to, w;
} adj[MAXM << 1];
void addedge(int from, int to, int w) {
  adj[++nume] = (Adj){head[from], to, w};
  head[from] = nume;
}
void link(int u, int v, int w) {
  addedge(u, v, w);
  addedge(v, u, w);
}
long long n, m;
long long p[60];
int main() {
  n = in();
  m = in();
  p[0] = 1;
  for (int i = 1; i <= 50; ++i) {
    p[i] = p[i - 1] * 2;
  }
  long long ans = 1e18;
  for (long long i = 1; i <= 1000000; ++i) {
    long long now = n - i * m;
    long long need = 0, needm = 0;
    for (int j = 50; j >= 0; --j) {
      if (p[j] > now) continue;
      ++need;
      now -= p[j];
      needm += p[j];
    }
    if (now != 0) continue;
    if (need <= i && i <= needm) {
      ans = min(ans, i);
      break;
    }
  }
  if (ans == 1000000000000000000)
    printf("-1");
  else
    printf("%lld\n", ans);
  return 0;
}

```