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>

long long g[210000] = {0};

long long p(long long n)
{
	long long ans = 0;
	long long i;
	for (i = 0; n >> i; i++)
	{
		if ((n >> i) & 1) ans++;
	}
	return ans;
}

long long f(long long n)
{
	long long ans = 0;
	while (n)
	{
		if (n < 10000) n = g[n];
		else n = n % p(n);
		ans++;
	}
	return ans;
}

int main(void)
{
	long long n, x;
	char str[210000];
	long long i, j, max, xi, r;

	scanf("%lld", &n);
	scanf("%s", str);

	x = 0;
	for (i = 0; i < n; i++)
	{
		x *= 2;
		x += str[i] - '0';
	}

	max = 1;
	for (i = 0; i < n; i++)
	{
		max *= 2;
	}
	max--;
	//printf("%lld\n", max);

	for (i = 10000; i > 0; i--)
	{
		g[i] = i % p(i);
		//printf("g[%lld] = %lld\n", i, g[i]);
	}

	r = (max + 1) / 2;
	for (i = 0; i < n; i++)
	{
		xi = (x ^ r);
		printf("%lld\n", f(xi));
		//printf("%lld\n", xi);
		r /= 2;
	}

	return 0;
}