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<stdio.h>
#include<stdlib.h>
#include<math.h>

int popcount(int m);
int makeans(long long int numi,int onei);

int main(void){
	int n,one=0;
	long long int num=0;
	scanf("%d",&n);
	int *x;
	if((x=malloc(n*sizeof(int)))==NULL){
		printf("error");
		return 0;
	}
	
	for(int i=0;i<n;i++){
		scanf("%1d",&x[i]);
		num=num*2+(long long int)x[i];
		if(x[i]){
			one++;
		}
	}
	
	int temp,one_i;
	long long int num_i;
	for(int i=0;i<n;i++){
		if(x[i]==0){
			num_i=num+(long long int)pow(2,n-i-1);
			one_i=one+1;
		}else{
			num_i=num-(long long int)pow(2,n-i-1);
			one_i=one-1;
		}
		//printf("%d %d",num_i,one_i);
		temp=makeans(num_i,one_i);
		printf("%d\n",temp);
	}
		
	free(x);
}

int makeans(long long int numi,int onei){
	//printf("(%d,%d)\n",numi,onei);
	int temp;
	if(onei==0){
		return 1;
	}else if((numi%onei)==0){
		return 1;
	}else{
		temp=makeans(numi%(long long int)onei,popcount(numi%onei));
		return temp+1;
	}
}

int popcount(int m){
	int cot=0;
	while(m!=0){
		cot+=(m%2);
		m=(m-(m%2))/2;
	}
	return cot;
}