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: Accepted)

#include <iostream>
#include <string>
#include <cmath>
using namespace std;

int main() {
  	int N;
  	cin >> N;
  	string X;
  	cin >> X;
  
  	int help[N + 1];
  	for (int i = 0; i < N + 1; i++) help[i] = 0;
  
  	for (int i = 0; i < N + 1; i++){
      	if (i == 0) continue;
      	int count = 0;
      	int a = 1;
      	int b = i;
      	while (a < i){
          	a *= 2;
        }
      	while (a != 0){
          	if (a > b) a /= 2;
          	else {
              	count++;
              	b -= a;
              	a /= 2;
            }
        }
      	help[i] = i % count;
    }
  
  	int dp[N + 1];
  	for (int i = 0; i < N + 1; i++){
      	if (i == 0){
          	dp[0] = 0;
          	continue;
        }
      	dp[i] = dp[help[i]] + 1;
    }
  
  	int one = 0; // 1の数
  	for (int i = 0; i < N; i++){
      	if (X[i] == '1') one++;
    }
  
  	int a = 1;
  	int second_pow_plus[N]; // 1が多い時の累積和
  	for (int i = 0; i < N; i++){
      	if (i >= 1) a *= 2;
      	a %= one + 1;
      	second_pow_plus[i] = a;
    }
  
  	a = 1;
  	int second_pow_minus[N]; // 1が少ない時の累積和
  	if (one >= 2) for (int i = 0; i < N; i++){
      	if (i >= 1) a *= 2;
      	a %= one - 1;
      	second_pow_minus[i] = a;
    }
  
  	int all_plus = 0; //1が多い時の元の数字
  	for (int i = 0; i < N; i++){
      	if (X[i] == '1'){
          	all_plus += second_pow_plus[N - i - 1];
          	all_plus %= one + 1;
        }
    }
  
  	int all_minus = 0; //1が少ない時の元の数字
  	if (one >= 2) for (int i = 0; i < N; i++){
      	if (X[i] == '1'){
          	all_minus += second_pow_minus[N - i - 1];
          	all_minus %= one - 1;
        }
    }
  
  	int res[N]; 
  	for (int i = 0; i < N; i++){
      	if (X[i] == '0'){
          	int aaaa = second_pow_plus[N - i - 1] + all_plus;
          	aaaa %= one + 1;
          	res[i] = dp[aaaa] + 1;
        }
      	if (X[i] == '1'){
          	if (one >= 2){
          	int aaaa = all_minus - second_pow_minus[N - i - 1];
          	aaaa += one - 1;
          	aaaa %= one - 1;
          	res[i] = dp[aaaa] + 1;
            }
          	else res[i] = 0;
        }
    }
  
  	for (int i = 0; i < N; i++){
      	cout << res[i] << endl;
    }
}