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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define SIZE	(200000)
#define UNIT	(32)		// unsigned long で保証される bit 長

// bits[len] の上位 32bit までを unisgned long 型にして戻す
unsigned long upper(unsigned long len, char bits[])
{
	unsigned long i, n = 0;

	for (i = 0; len-- > 0 && i < UNIT; i++) {
		n = 2 * n + bits[i];
	}
	return n;
}

// bits[len] を p で割った余りに書き換える
void mod(int len, char *bits, unsigned long p) {
	while (len > 0) {
		int i, u = (len > UNIT) ? UNIT : len;
		unsigned long n = upper(u, bits) % p;

		for (i = 0; i < u; i++) {
			bits[u - 1 - i] = (char)(n % 2);
			n /= 2;
		}
		// 10^5 < 2^17 なので p は高々 17 bit
		// ゆえに n も高々 17 bit
		len -= UNIT - 17;
		bits += UNIT - 17;
	}
}

int popcount(unsigned long len, char bits[]) {
	unsigned long i, count = 0;

	for (i = 0; i < len; i++) {
		if (bits[i])
			count++;
	}
	return count;
}

int f(unsigned long len, char bits[])
{
	char buf[SIZE], *b = buf;
	unsigned long p;
	int count = 0;

	memcpy(b, bits, len);
	while (1) {
		while (len > 0 && b[0] == 0) {
			b++;
			len--;
		}
		p = popcount(len, b);
		if (p == 0)
			break;
		mod(len, b, p);
		count++;
	}
	return count;
}

int main()
{
	unsigned long i, len;
	char bits[SIZE + 1];

	scanf("%d%s", &len, bits);
	for (i = 0; i < len; i++)
		bits[i] = (bits[i] == '1');

	for (i = 0; i < len; i++) {
		bits[i] = !bits[i];
		printf("%d\n", f(len, bits));
		bits[i] = !bits[i];
	}
	return 0;
}