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>
#define N 200005
int n, m, i, j, k, a, b, x, y, p, q;
int u[N], v[N];
char s[N];
void f(int k, int n){
	if(n == 0) printf("%d\n", k);
	else{
		int m=n, c=0;
		while(m){
			c += m&1;
			m >>= 1;
		}
		f(k+1, n%c);
	}
}
int main(){
	scanf("%d%s", &n, s+1);
	for(i=1; i<=n; i++){
		a += s[i] & 1;
	}
	b = a + 1;
	if(a) a--;
	u[n] = v[n] = 1;
	for(p=q=1, i=n; i>=1; i--){
		if(s[i] & 1){
			x = (x+p) % a;
			y = (y+q) % b;
		}
		u[i-1] = p = (p+p) % a;
		v[i-1] = q = (q+q) % b;
	}
	for(i=1; i<=n; i++){
		if(s[i]&1) f(1, (x-u[i]+a)%a);
		else f(1, (y+v[i])%b);
	}
	return 0;
}