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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define MIN(a,b) ((a)>(b)? (b): (a))
#define MAX(a,b) ((a)<(b)? (b): (a))

const long long INF = 1LL << 60;
typedef unsigned long long ll;
const long long MOD = 1000000000 + 7;

int strbin2i(const std::string &s)
{

    int out = 0;

    for (int i = 0, size = s.size(); i < size; ++i)
    {

        out *= 2;

        out += ((int)s[i] == 49) ? 1 : 0;
    }

    return out;
}

int main(){
    int n;
    cin >> n;
    string s;
    cin >> s;
    int bit = strbin2i(s);
    int bit_tmp;
    for (int i = n-1; i >= 0; i--)
    {
        bit_tmp = bit;
        if (bit_tmp & (1 << i))
        {
            bit_tmp &= ~(1 << i);
        }
        else
        {
            bit_tmp |= (1 << i);
        }
        int cnt,cnt_pop ;
        cnt = 0;
        while (bit_tmp > 0)
        {
            cnt_pop = __builtin_popcount(bit_tmp);
            bit_tmp = bit_tmp % cnt_pop;
            cnt++;
        }
        cout << cnt << endl;
    }
    return 0;
}