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;

vector<long long> f(200001);

long long power(long long x, long long y, long long mod)
{
    long long ret = 1;
    while(y > 0)
    {
        if (y & 1)
        {
            ret = (ret * x) % mod;
        }
        x = (x * x) % mod;
        y >>= 1;
    }
    return ret;
}

long long func(long long count, long long x)
{
    if (x % count == 0)
    {
        return 1;
    }
    else
    {
        return 1 + f[x % count];
    }
}

int main(void)
{
    for (long long i = 1; i < 200001; i++)
    {
        long long count = 0;
        long long X = i;
        while(X > 0)
        {
            if(X & 1)
            {
                count ++;
            }
            X >>= 1;
        }
        f[i] = func(count, i);
    }
    long long N;
    cin >> N;
    string X;
    cin >> X;
    long long count = 0;
    for (long long i = 0; i < X.size(); i++)
    {
        if (X[i] == '1')
        {
            count++;
        }
    }
    long long num1 = 0;
    for (long long j = 0; j < X.size(); j++)
    {
        if (X[j] == '1')
        {
            num1 += power(2,X.size()-1-j, count-1);
        }
    }
    long long num0 = 0;
    for (long long j = 0; j < X.size(); j++)
    {
        if (X[j] == '1')
        {
            num0 += power(2,X.size()-1-j, count+1);
        }
    }
    for (long long i = 0; i < X.size(); i++)
    {
        long long answer;
        if (X[i] == '1')
        {
            answer = func(count - 1, num1 - power(2,X.size()-1-i, count-1));
        }
        else
        {
            
            answer = func(count + 1, num0 + power(2,X.size()-1-i, count+1));
        }
        cout << answer << endl;
    }
}