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 maxn = 1e5 + 6;
const long long int MOD = 1e9 + 7;
vector<int> prim(1000005, 1);
int fact[maxn];
long long int binomialCoeff(long long int n, long long int k) {
  long long int res = 1;
  if (k > n - k) k = n - k;
  for (long long int i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
bool isVowel(char ch) {
  if (ch == 'a' || ch == 'i' || ch == 'e' || ch == 'u' || ch == 'o') {
    return true;
  }
  return false;
}
long long int power(long long int x, long long int i, long long int mod) {
  long long int ans = 1;
  while (i > 0) {
    if (i & 1) ans = (ans * x) % mod;
    i >>= 1;
    x = (x * x) % mod;
  }
  return ans;
}
long long int modInverse(long long int x, long long int mod) {
  return power(x, mod - 2, mod);
}
int nCr(int n, int r) {
  if (n < r) {
    return 0;
  }
  return (((fact[n] * modInverse(fact[n - r], MOD)) % MOD) *
          modInverse(fact[r], MOD)) %
         MOD;
}
long long int power(int x, unsigned int y) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else
    return x * temp * temp;
}
void erath(int n) {
  prim[1] = 0;
  prim[0] = 0;
  prim[2] = 1;
  for (int i = 2; i * i <= n; i++) {
    if (prim[i]) {
      for (int j = i * i; j <= n; j += i) {
        prim[j] = 0;
      }
    }
  }
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a / gcd(a, b)) * b;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long int n, m, k;
int getmid(int l, int r) { return (l + (r - l) / 2); }
struct segtree {
  int siz;
  vector<pair<long long int, long long int> > sums;
  void init(int n) {
    siz = 1;
    while (siz < n) siz *= 2;
    sums.assign(2 * siz, {LONG_MAX, 0});
  }
  void set(int i, int v, int cur, int l, int r) {
    if (r - l == 1) {
      sums[cur].first = v;
      sums[cur].second = 1;
      return;
    }
    int m = getmid(l, r);
    if (i < m) {
      set(i, v, 2 * cur + 1, l, m);
    } else
      set(i, v, 2 * cur + 2, m, r);
    sums[cur].first = min(sums[2 * cur + 1].first, sums[2 * cur + 2].first);
    if (sums[2 * cur + 1].first == sums[cur].first &&
        sums[2 * cur + 2].first == sums[cur].first) {
      sums[cur].second = sums[cur * 2 + 1].second + sums[cur * 2 + 2].second;
    } else if (sums[2 * cur + 1].first == sums[cur].first) {
      sums[cur].second = sums[cur * 2 + 1].second;
    } else {
      sums[cur].second = sums[cur * 2 + 2].second;
    }
  }
  void set(int i, int v) { set(i, v, 0, 0, siz); }
  pair<long long int, long long int> sum(int l, int r, int cur, int lx,
                                         int rx) {
    if (lx >= r || l >= rx) return {LONG_MAX, 0};
    if (lx >= l && rx <= r) return {sums[cur].first, sums[cur].second};
    int mid = getmid(lx, rx);
    pair<long long int, long long int> s1 = sum(l, r, 2 * cur + 1, lx, mid),
                                       s2 = sum(l, r, 2 * cur + 2, mid, rx);
    if (s1.first < s2.first) {
      return s1;
    } else if (s1.first > s2.first)
      return s2;
    else
      return {s1.first, s1.second + s2.second};
  }
  pair<long long int, long long int> sum(int l, int r) {
    return sum(l, r, 0, 0, siz);
  }
};
int nCrModp(int n, int r, int p) {
  if (r > n - r) r = n - r;
  int C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = min(i, r); j > 0; j--) C[j] = (C[j] + C[j - 1]) % p;
  }
  return C[r];
}
void ans(int top) {
  cin >> n >> k;
  int ans = INT_MAX;
  for (int i = 1; i < 10000; i++) {
    m = n - i * k;
    if (m < i) continue;
    int cur = 0;
    while (m > 0) {
      if (m == 1) {
        cur++;
        break;
      }
      int minu = log2(m);
      m -= pow(2, minu);
      cur++;
    }
    if (cur <= i) ans = min(ans, i);
  }
  if (ans == INT_MAX) {
    cout << "-1\n";
    return;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t = 1, co = 1;
  while (t--) {
    ans(co);
    co++;
  }
  return 0;
}
