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 <stdio.h>

int main()
{
	int N;
	char X[200001];
	scanf("%d", &N);
	scanf("%s", X);
	
	int i, sum = 0;
	for (i = 0; X[i] != 0; i++) sum += X[i] - '0';
	if (sum == 0) {
		for (i = 0; i < N; i++) printf("1\n");
		fflush(stdout);
		return 0;
	} else if (sum == 1) {
		if (X[N-1] == '0') {
			for (i = 0; i < N - 1; i++) {
				if (X[i] == '0') printf("1\n");
				else printf("0\n");
			}
			printf("2\n");
		} else {
			for (i = 0; i < N - 1; i++) printf("2\n");
			printf("0\n");
		}
		fflush(stdout);
		return 0;
	}
	
	int res[2][200001] = {}, total[2] = {X[N-1] - '0', X[N-1] - '0'}, Y, Z, pop, count;
	for (i = N - 2, res[0][N-1] = 1, res[1][N-1] = 1; i >= 0; i--) {
		res[0][i] = res[0][i+1] * 2 % (sum + 1);
		res[1][i] = res[1][i+1] * 2 % (sum - 1);
		total[0] = (total[0] + (X[i] - '0') * res[0][i]) % (sum + 1);
		total[1] = (total[1] + (X[i] - '0') * res[1][i]) % (sum - 1);
	}
	for (i = 0; i < N; i++) {
		if (X[i] == '0') Y = (total[0] + res[0][i]) % (sum + 1);
		else Y = (total[1] - res[1][i] + sum - 1) % (sum - 1);
		count = 1;
		while (Y > 0) {
			for (Z = Y, pop = 0; Z > 0; Z >>= 1) pop += Z % 2;
			Y %= pop;
			count++;
		}
		printf("%d\n", count);
	}
	fflush(stdout);
	return 0;
}