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<bits/stdc++.h>

using namespace std;

typedef long long ll;

void cldflm()
{
	std::ios_base::sync_with_stdio(0);
	cin.tie(0);

#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w" , stdout);
#endif
}

int popcount(int num);

int main()
{

	cldflm();

	int n;
	string str;
	cin >> n >> str;
	int i = 0;

	while (n--)
	{
		int cnt = 0;
		string tempstr;
		tempstr = str;
		if (tempstr[i] == '0')
			tempstr[i] = '1';
		else
			tempstr[i] = '0';
		ll number = stoi(tempstr, 0, 2);

		while (number > 0)
		{
			int kk = popcount(number);
			number = number % kk;
			cnt++;
		}

		cout << cnt << "\n";
		i++;

	}


	return 0;
}

int popcount(int num)
{
	int count = 0;
	while (num != 0)
	{
		num = num & (num - 1);
		count++;
	}
	return count;
}