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 int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<n;i++)
#define rev(i,n) for(int i=n-1;i>=0;i--)
#define all(v) v.begin(),v.end()
#define P pair<int,int>
#define len(s) (int)s.size()

template<class T> inline bool chmin(T &a, T b){
	if(a>b){a=b;return true;}
	return false;
}
template<class T> inline bool chmax(T &a, T b){
	if(a<b){a=b;return true;}
	return false;
}
constexpr int mod = 1e9+7;
constexpr int inf = 3e18;

int ans[200005];
int f(int x){
  	if(x==0)return 0;
	if(ans[x])return ans[x];
	return ans[x]=f(x%__builtin_popcount(x))+1;
}
int div1[200005],div2[200005];
signed main(){
	int N;std::string S;
	std::cin>>N>>S;
	int cnt=0,now1=1,now2=1;
	rep(i,N){
		if(S[i]=='1')cnt++;
	}
	int ans1=0,ans2=0;
	rev(i,N){
		div1[N-1-i]=now1;
		div2[N-1-i]=now2;
		if(S[i]=='1'){
			ans1+=now1;ans2+=now2;
			ans1%=cnt+1;
			if(cnt!=1)ans2%=cnt-1;
		}
		now1*=2;
		now1%=cnt+1;
		if(cnt!=1){
			now2*=2;now2%=cnt-1;
		}
	}
	rep(i,N){
		if(cnt==1&&S[i]=='1')cout<<"0\n";
		else if(S[i]=='1'){
			int j=(ans2-div2[N-i-1]+cnt-1)%(cnt-1);
			cout<<f(j)+1<<"\n";
		}else {
			int j=(ans1+div1[N-i-1])%(cnt+1);
			cout<<f(j)+1<<"\n";
		}
	}
}