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 2x+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=20−9, 7=24−9, 1015=210−9).
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 20+21+22.
And if p=−9 we can represent 7 as one number (24−9).
## 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≤109, −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
```cpp
24 0
2
24 1
3
24 -1
4
4 -7
2
1 1
-1
0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24=(24+0)+(23+0).
In the second sample case, we can represent 24=(24+1)+(22+1)+(20+1).
In the third sample case, we can represent 24=(24−1)+(22−1)+(22−1)+(22−1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4=(24−7)+(21−7). Note that the second summand is negative, which is allowed.
In the fifth sample case, no representation is possible.
找规律,先枚举有多少项加起来,然后拆分数位,如果1的个数小于枚举的项数,那就是可行的。
/*
`-._: .:' `::: .:\ |\__/| /:: .:' `::: .:.-'
\ : \ | :: | / : /
\ :: . `-_______/ :: \_______-' . :: . /
| : :: ::' : :: ::' ::: ::' :: ::' : :: : |
| ;:: ;:: ;:: ;:: ;:: |
| .:' `::: .:' `::: .:' ::: .:' `::: .:' `: |
/ : : : : : \
/______::_____ :: . :: . :: _____._::____\
`----._:: ::' : :: ::' _.----'
`--. ;:: .--'
`-. .:' .-'
\ /
\ /
\/
*/
#include
//#include
//#include
//#include
//#include
//#include
//#include
#define read(x) scanf("%lld",&x)
#define re(n) for(int i = 0 ; i < n ; i ++)
#define rev(n) for(int i = n-1 ; i >= 0 ; i --)
#define fill(x,y) memset(x,y,sizeof(x))
const int N=1e6+10;
const int INF=0x3f3f3f3f;
typedef long long ll;
using namespace std;
int main()
{
ll n,m,i,j,k,t = 1;
for(int cas = 1 ; cas <= t ; cas ++)
{
cin>>n>>m;
ll ans = 0;
if(n <= m)
cout<<"-1"<<endl;
else{
while(1)
{
ans ++;
ll tmp = n-ans*m;
if(tmp < ans)
{
cout<<"-1"<<endl;
return 0;
}
ll tmp1 = 0;
while(tmp)
{
tmp1 += tmp%2;
tmp /= 2;
}
if(tmp1 <= ans)
break;
}
cout<<ans<<endl;
}
//cout<<"Case "<
}
return 0;
}