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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define debug(x) cout<<x
#define pb push_back
#include<string.h>
ll modex(ll a, ll b, ll mod){
	ll ans=1;
	while(b){
		if(b%2) ans=(ans*a)%mod;
		a=(a*a)%mod;
		b/=2;
	}
	return ans;
}
ll count(ll val){
	ll cnt=0;
	while(val){
		if(val%2) cnt++;
		val/=2;
	}
	return cnt;
}
ll fun(ll val){
	if(val==0) return 0;
	return (1+fun(val%count(val)));
}
int main(){
	ios_base::sync_with_stdio(false); 
    cin.tie(NULL);                    
    cout.tie(NULL);
    ll T=1;
    //cin>>T;
    while(T--){
    	ll n;
    	cin>>n;
    	string str;
    	cin>>str;
    	ll val=0,val1=0,val2=0,cnt=0;
    	for(int i=0;i<n;i++) if(str[i]=='1') cnt++;
    	for(int i=0;i<n;i++){
    		val=(val+(int)(str[i]-'0')*modex(2,n-1-i,cnt))%cnt;
    		val1=(val1+(int)(str[i]-'0')*modex(2,n-1-i,cnt+1))%(cnt+1);
    		if(cnt>1)val2=(val2+(int)(str[i]-'0')*modex(2,n-1-i,cnt-1))%(cnt-1);
    	}
    	for(int i=0;i<n;i++){
    		if(str[i]=='1'){
    			if((cnt-1)==0) {
    				cout<<"0"<<endl;
    				continue;
    			}
    			ll value=((((val2-modex(2,n-1-i,cnt-1))%(cnt-1))+(cnt-1))%(cnt-1));
    			cout<<(1+fun(value))<<endl;
    		}
    		else{
    			ll value=((val1+modex(2,n-1-i,cnt+1))+cnt+1)%(cnt+1);
    			cout<<(1+fun(value))<<endl;
    		}
    	}
    }
   	return 0; 
}