Problem:
Score : 400 points

Problem Statement

Let \mathrm{popcount}(n) be the number of 1s in the binary representation of n.
For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.

Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)

For example, when n=7, it becomes 0 after two operations, as follows:

\mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.

\mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.

You are given an integer X with N digits in binary.
For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted.
Find f(X_1), f(X_2), \ldots, f(X_N).

Constraints

1 \leq N \leq 2 \times 10^5

X is an integer with N digits in binary, possibly with leading zeros.

Input

Input is given from Standard Input in the following format:

N
X

Output

Print N lines. The i-th line should contain the value f(X_i).

Sample Input 1

3
011

Sample Output 1

2
1
1

X_1 = 7, which will change as follows: 7 \rightarrow 1 \rightarrow 0. Thus, f(7) = 2.

X_2 = 1, which will change as follows: 1 \rightarrow 0. Thus, f(1) = 1.

X_3 = 2, which will change as follows: 2 \rightarrow 0. Thus, f(2) = 1.

Sample Input 2

23
00110111001011011001110

Sample Output 2

2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define ll long long
using namespace std;
char a[200005];
ll quickpow(ll a,ll b,ll mod)
{
	ll ans=1,base=a;
	while(b)
	{
		if(b&1)
		ans=(ans*base)%mod;
		ans=(ans*ans)%mod;
		b>>=1;
	}return ans%mod;
 } 
ll cal(ll res)
{
		ll num=1;
				while(res)
				{
					res=res%__builtin_popcount(res);
					num++;
				}
				return num;
}
int main()
{
	ll n,sum=0;ll res1=0,res2=0,res=0;cin>>n;getchar();
	for(ll i=1;i<=n;++i)
	{
		char c;
		c=getchar();a[i]=c;
		if(c=='1') sum++;
	}
	if(sum>1)
	{
		for(ll i=1;i<=n;++i)
		{
			if(a[i]=='1') 
			{
			res1=(res1+quickpow(2,n-i,sum-1))%(sum-1);
			res2=(res2+quickpow(2,n-i,sum+1))%(sum-1);	
			}
		}
		for(ll i=1;i<=n;++i)
		{
			if(a[i]=='1')
			{
				res=(res1+sum-1-quickpow(2,n-i,sum-1))%(sum-1);
				cout<<cal(res)<<endl;
			}
			else 
			{
			res=(res2+quickpow(2,n-i,sum+1))%(sum+1);
			cout<<cal(res)<<endl;	
			}
		}
	}else if(sum==1)
	{
	for(int i=1;i<=n;++i)
		{
			if(a[i]=='1') 
			{
			res2=(res2+quickpow(2,n-i,sum+1))%(sum+1);	
			}
		}
		for(int i=1;i<=n;++i)
		{
			if(a[i]=='0') 
			{
				res=(res2+quickpow(2,n-i,sum+1))%(sum+1);
				cout<<cal(res)<<endl;
			}
			else cout<<'0'<<endl;
		}
	 } 
	 else {
	 	for(int i=1;i<=n;++i)
	 	cout<<'1'<<endl;
	 }
 }