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 optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
using i32 = int;
using i64 = long long;
using u8 = unsigned char;
using u32 = unsigned;
using u64 = unsigned long long;
using f64 = double;
using f80 = long double;
long long power(long long a, long long b, long long p) {
  if (!b) return 1;
  long long t = power(a, b / 2, p);
  t = t * t % p;
  if (b & 1) t = t * a % p;
  return t;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long px, py;
  long long d = exgcd(b, a % b, px, py);
  x = py;
  y = px - a / b * py;
  return d;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <class T>
inline void freshmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <class T>
inline void freshmax(T &a, const T &b) {
  if (a < b) a = b;
}
template <class T>
void print(const T &a) {
  for (auto x : a) printf("%d ", x);
  puts("");
}
const int MAXN = 500010;
const int MAXK = 30;
const int INF = 1000000000;
const int B = 31;
const int MOD = 1000000007;
const double pi = 3.1415926535897932384626433;
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
int days[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool leap(int y) { return y % 4 == 0 && y % 100 != 0 || y % 400 == 0; }
int n, m;
int a[MAXN], b[MAXN];
void solve() {
  int n, p;
  cin >> n >> p;
  for (int k = 1; k <= 1000000; ++k) {
    n -= p;
    if (n <= 0) break;
    int m = __builtin_popcount(n);
    if (m <= k && k <= n) {
      printf("%d\n", k);
      return;
    }
  }
  puts("-1");
}
int main() {
  int T;
  T = 1;
  while (T--) solve();
  return 0;
}
