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)

#pragma GCC optimize("O3")
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<functional>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
#define MOD 1000000007
#define f(i,n) for(int i=0;i<(int)(n);i++)
#define N 200010


int main() {
	char s[N];
	f(i, N)s[i] = 0;
	int n;
	int ans[N];
	int fians[N];
	int a[N];
	int b[N];
	int c[N];
	int aa, bb, cc;
	int ka, kb, kc;
	int x, y, z;
	ans[0] = 0;
	f(i, N-1) {
		x = 0;
		y = i + 1;
		while (y > 0) {
			if (y % 2 == 1)x++;
			y /= 2;
		}
		x = (i + 1) % x;
		ans[i + 1] = ans[x] + 1;
	}
	scanf("%d", &n);
	scanf("%s", s);
	ka = 0;
	f(i, n) {
		if (s[i] == '1')ka++;
	}
	kb = ka - 1;
	kc = ka + 1;
	if (kb > 0) {
		x = 1 % kb;
		bb = 0;
		f(i, n) {
			b[i] = x;
			if (s[n - i - 1] == '1')bb = (bb + x) % kb;
			x = (x * 2) % kb;
		}
	}
	if (kc > 0) {
		x = 1 % kc;
		cc = 0;
		f(i, n) {
			c[i] = x;
			if (s[n - i - 1] == '1')cc = (cc + x) % kc;
			x = (x * 2) % kc;
		}
	}
	f(i, n) {
		if (s[n-i-1] == '1') {
			if (kb <= 0)fians[n - i - 1] = 0;
			else {
				x = bb + (kb - b[i]);
				x = x % kb;
				fians[n - i - 1] = ans[x] + 1;
			}

		}
		else {
			x = cc + c[i];
			x = x % kc;
			fians[n - i - 1] = ans[x] + 1;
		}

	}
	f(i,n)printf("%d\n", fians[i]);


	return 0;
}