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: Runtime Error)

#include<bits/stdc++.h>
#define F first
#define S second
#define ll long long
#define MOD 1000000007
#define fast ios_base::sync_with_stdio(false)
#define itt cin.tie(0)
#define now cout.tie(0)
#define N 200010
using namespace std;
ll n,nn[2],md[N][2];
string s;
ll f(ll v,ll nd,ll sd,ll sx){
	ll ret=1,x=(nn[nd]+sd+sx*md[v][nd])%sd;
	while(x!=0)
	{
		x=x%__builtin_popcount(x);
		ret++;
	}
	return ret;
}
int main(){
	ll i,bb=0;
	fast;itt;now;
	cin>>n>>s;
	for(i=0;i<n;i++)
	{
		if(s[i]=='1')
		{
			bb++;
		}
	}
	md[0][0]=1%(bb+1);
	md[0][1]=1%(bb-1);
	for(i=1;i<=n;i++)
	{
		md[i][0]=(md[i-1][0]*2)%(bb+1);
		md[i][1]=(md[i-1][1]*2)%(bb-1);
	}
	nn[0]=nn[1]=0;
	for(i=0;i<n;i++)
	{
		nn[0]=(nn[0]*2+(s[i]-'0'))%(bb+1);
		nn[1]=(nn[1]*2+(s[i]-'0'))%(bb-1);
	}
	for(i=0;i<n;i++)
	{
		if(s[i]=='0')
		{
			cout<<f(n-i-1,0,bb+1,1)<<endl;
		}
		else
		{
			cout<<f(n-i-1,1,bb-1,-1)<<endl;
		}
	}
	return 0;
}