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 <iostream>
#include <string>

using namespace std;


const int two20 = 1048576;


int main()
{
	int n;
	string s;

	cin >> n >> s;

	int divnum = 0;
	for (int i = 0; i < n; i++) {
		if (s.substr(i, 1) == "1") divnum++;
	}

	int aaa[100000];
	for (int i = 0; i < 100000; i++) aaa[i] = 0;

	bool cont = true;

	for (int j = 0; cont; j++) {
		int kakeru = 1;
		for (int i = 0; i < 20; i++) {

			if (i + j * 20 >= s.length()) {
				cont = false;
				break;
			}

			if (s.substr(s.length() - i - 1 - j * 20, 1) == "1") aaa[j] += kakeru;
			kakeru *= 2;
			
		}
	}

	int start_1 = divnum + 1;
	int get_1 = 0;

	int baseKakeru = 1;

	for (int i = 0; i * 20 < s.length(); i++) {

		get_1 += (aaa[i] % start_1 + baseKakeru) % start_1;
		
		baseKakeru = (baseKakeru * two20) % start_1;
	}


	int start_2 = divnum - 1;
	int get_2 = 0;

	baseKakeru = 1;

	for (int i = 0; i * 20 < s.length(); i++) {

		get_2 += (aaa[i] % start_2 + baseKakeru) % start_2;

		baseKakeru = (baseKakeru * two20) % start_2;
	}





	//問題を解く

	for (int qqqqq = 0; qqqqq < n; qqqqq++) {
		int answer = 1;

		int rest;

		//get_1をつかう
		if (s.substr(qqqqq, 1) == "0") {
			
			int a = 1;
			for (int i = 0; i * 20 < qqqqq; i++) {
				a = (a * two20) % start_1;
			}

			for (int i = 0; i < qqqqq % 20; i++) {
				a = (a * 2) % start_1;
			}

			rest = (get_1 + a) % start_1;


		}

		else {


			int a = 1;
			for (int i = 0; i * 20 < qqqqq; i++) {
				a = (a * two20) % start_2;
			}

			for (int i = 0; i < qqqqq % 20; i++) {
				a = (a * 2) % start_2;
			}

			rest = (get_2 + start_2 - a) % start_2;
		}



		//ここからちゃんと求める

		while (rest > 0) {
			int rests = rest;
			int divs = 0;
			for (int i = 0; rests > 0; i++) {
				divs += rests % 2;
				rests /= 2;
			}

			rest = rest % divs;
			answer++;
		}

		cout << answer << endl;
	}


}