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 int MOD = 1000000007;
const long long BIG = 1446803456761533460LL;
const int Big = 336860180;
const long long int INF = LONG_LONG_MAX;
const long long int adj4[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const long long int adj8[8][2] = {{0, 1},   {0, -1}, {1, 0},  {-1, 0},
                                  {-1, -1}, {1, -1}, {-1, 1}, {1, 1}};
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int powMod(long long int a, long long int b) {
  long long int n = 1;
  long long int p = a;
  while (b > 0) {
    if (b % 2 == 1) {
      n *= p;
      n %= MOD;
    }
    p *= p;
    p %= MOD;
    b /= 2;
  }
  return n;
}
long long int modularInverse(long long int a) { return powMod(a, MOD - 2); }
stringstream sss;
const long long int maxn = 100010;
void MAIN() {
  long long int n, p;
  cin >> n >> p;
  for (long long int i = (1); i < (1000); ++i) {
    long long int v = n - i * p;
    if (v >= i && __builtin_popcount(v) <= i) {
      cout << (i) << '\n';
      return;
    }
  }
  cout << (-1) << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  sss << R"(
24 1
    )";
  MAIN();
  return 0;
}
