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: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <bitset>
#include <map>
using namespace std;
std::map<int, long long> mp1;
std::map<int, long long> mp2;
unsigned long ans1 = 0;
unsigned long ans2 = 0;
 
long long pow(long long n, long long p, long long m)
{
    if(m == 0) /*{printf("a"); exit(0);}*/ return 0;
    if(p == 0) return (1 % m);
    long long ret = 1;
    for (; p > 0; p >>= 1, n = n * n % m){
		if (p%2 == 1){
			ret = ret * n % m;
		}
	}
	return ret;
}

unsigned long f(std::bitset<200000> bs, long long index, long long N)
{
    long long ret = -1;
    long long count = 0;
    
    // 初回だけmpを使って計算
    int hibit = bs.count();
    if(bs[index] == 0){
        ret = ans1 + mp1[index];
        hibit++;
    }else{
        ret = ans2 - mp2[index];
        hibit--;
        if(hibit <= 0) return count;
        if(ret < 0) ret += hibit;
    }
    if(ret >= hibit){
        ret = ret % hibit;
    }
    count++;
    //std::cout << "hibit:" << hibit << ", ret:" << ret << std::endl;

    while(ret != 0){
        std::bitset<200000> tmpbs(ret);
        ret = ret % tmpbs.count();
        count++;
    }

    // nが0になるまでの操作回数を返す
    return count;
}
 
int main()
{
    long long N;
    std::cin >> N;
    std::string X;
    std::cin >> X;
 
    // 前準備
    std::bitset<200000> bs(X);
    const int count = bs.count();

    // 制約から1<=Nなので、count=0のケースは考慮不要
    for(long long i=N-1; i>=0; i--){
        mp1[i] = pow(2, i, count+1);
        mp2[i] = pow(2, i, count-1);
    }
    for(long long i=N-1; i>=0; i--){
       if(bs[i] == 1){
            ans1 += pow(2, i, count+1);
            ans2 += pow(2, i, count-1);
       }
    }
 
    // ここから本処理    
    for(long long i=N-1; i>=0; i--){
        unsigned long result = f(bs, i, N);
        std::cout << result << std::endl;
    }
}