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>
long long int modpow(long long int a, long long int n, long long int b)
{
	long long int res = 1;
	while (n > 0)
	{
		if (n % 2 > 0)
			res = res * a % b;
		a = a * a % b;
		n /= 2;
	}
	return res;
}
int main()
{
	long long int n;
	scanf("%lld", &n);
	long long int i;
	long long  int x[200005];
	char s[200005];
	scanf("%s", s);
	for (i = 0; i < n; i++)
		x[i] = s[i] - '0';
	long long int c1 = 0;
	for (i = 0; i < n; i++)
		if (x[i] == 1)
			c1++;
	c1--;
	long long int f1 = 0;
	if (c1 > 0)
		for (i = 0; i < n; i++)
			f1 = (2 * f1 + x[i]) % c1;
	long long int f0 = 0;
	long long int c0 = c1 + 2;
	for (i = 0; i < n; i++)
		f0 = (2 * f0 + x[i]) % c0;
	long long int a, ans, c, d;
	for (i = 0; i < n; i++)
	{
		if (x[i] == 0)
		{
			a = (f0 + modpow(2, n - 1 - i, c0)) % c0;
			ans = 1;
			while (a > 0)
			{
				c = 0;
				d = a;
				while (d > 0)
				{
					c += d % 2;
					d /= 2;
				}
				ans++;
				a %= c;
			}
			printf("%lld\n", ans);
		}
		else
		{
			if (c1 == 0)
				printf("0\n");
			else
			{
				a = (f1 - modpow(2, n - 1 - i, c1) + c1) % c1;
				ans = 1;
				while (a > 0)
				{
					c = 0;
					d = a;
					while (d > 0)
					{
						c += d % 2;
						d /= 2;
					}
					ans++;
					a %= c;
				}
				printf("%lld\n", ans);
			}
		}
	}
	return 0;
}