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

long num(char s[],long n){
    long sum=0,j=0;
    for(long i=n-1;i>=0;i--){
        sum+=(pow(2,j))*(s[i]-48);
        //printf("sumis%ld %ld\n",sum,s[i]);
        j++;
    }
  return sum;
}

long bin(long n){
    long cnt=0,a;
     for(long i=0;n>0;i++){
         a=n%2;
         if(a==1) cnt++;
         n/=2;
     }
     return cnt;

}

int main() {
long n,i;
scanf("%ld",&n);
char s[100];
scanf("%s",s);
//printf("%ld %s",n,s);


for(i=0;i<n;i++){
    if(s[i]=='0') s[i]='1';
    else s[i]='0';
    //printf("%s\n",s);
    long r=num(s,n),cnt=0,k;
   // printf("r is %ld\n",num(s,n));
    
    while(r!=0){
        k=bin(r);
       // printf("kis%ld\n",k);
        cnt++;
        r=r%k;
    }
    printf("%ld\n",cnt);
    if(s[i]=='0') s[i]='1';
    else s[i]='0';
    
}

    return 0;
}