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 int INF = 1000000001;
const long long int INFLL = 1000000000000000001;
const long double EPS = 1e-6;
unordered_map<long long int, int> L, R;
unordered_map<long long int, int> Gen(const vector<long long int>& T) {
  int n = T.size();
  unordered_map<long long int, int> result;
  for (int i = 0; i < (1 << n); i++) {
    long long int s = 0;
    int cnt = 0;
    for (int j = 0; j < n; j++)
      if (i & (1 << j)) {
        s += T[j];
        cnt++;
      }
    if (result.count(s) == 0) result[s] = INF;
    result[s] = min(result[s], cnt);
  }
  return result;
}
int Greedy(long long int n) { return __builtin_popcountll(n); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, p;
  cin >> n >> p;
  vector<long long int> l, r;
  for (int i = 0; i < 16; i++) l.push_back((1 << i) + p);
  for (int i = 16; i < 31; i++) r.push_back((1 << i) + p);
  L = Gen(l);
  R = Gen(r);
  int best = INF;
  for (const pair<long long int, int>& x : L) {
    long long int y = n - x.first;
    if (R.count(y) == 0) continue;
    int rr = x.second + R[y];
    best = min(best, rr);
  }
  for (int i = 0; i <= min(100000, best); i++) {
    long long int amount = n - p * i;
    if (amount < i) continue;
    int denoms = Greedy(amount);
    if (denoms > i) continue;
    best = min(best, max(i, denoms));
  }
  if (best == INF)
    cout << "-1\n";
  else
    cout << best;
  return 0;
}
