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 <string.h>

long pop[200001];

void popcount(char *x) {
   long i,w,r,rr,ro,ans;

   for (w=i=0;x[i]!='\0';i++) 
      if (x[i]=='1') w++;
   for (r=i=0;x[i]!='\0';i++) {
      r*=2;
      if (x[i]=='1') r++;
      r%=w;
   }
   if (pop[r]<0) {
      ro=r;
      for (ans=1;r>0;ans++) {
         if (pop[r]>0) {
            ans+=pop[r];
            break;
         }
         for (w=0,rr=r;rr>0;rr/=2)
            if (rr%2==1) w++;
         r%=w;
      }
      pop[ro]=ans;
      printf("%ld\n",ans);
   } else
      printf("%ld\n",pop[r]);
}

int main(){
   int n,i;
   char x[200001],xx[200001];

   scanf("%d",&n);
   scanf("%s",x);
   for (i=0;i<200001;i++) pop[i]=-1;
   for (i=0;i<n;i++) {
      strcpy(xx,x);
      if (xx[i]=='0') xx[i]='1';
      else            xx[i]='0';
      popcount(xx);
   }
   return 0;
}