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<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define For(i,n,k) for(int i=(n);i<(k);i++)
#define ALL(a)  (a).begin(),(a).end()
ll ans=0;
void Main(){
    int n;
    string s;
    cin >> n >> s;
    vector<int> vec(n),ans(n,1);
    int t = count(ALL(s),'1');
    if(t==0){
        For(i,0,n){
            cout << ans[i] << endl;
        }
        return;
    }
    else if(t==1){
        if(s[n-1] == '1'){
            For(i,0,n-1){
                cout << 2 << endl;
            }
            cout << 0 << endl;
        }
        else{
            For(i,0,n-1){
                cout << 1 - (s[i] == '1') << endl;
            }
            cout << 2 << endl;
        }
        return;
    }
    int tp=0,tm=0;
    For(i,0,n){
        tp = tp * 2 + (s[i] == '1');
        tm = tm * 2 + (s[i] == '1');
        tp %= (t+1);
        tm %= (t-1);
    }
    reverse(ALL(s));
    int pp=1,pm=1;
    For(i,0,n){
        if(s[i] == '1'){
            vec[i] = ((tm - pm)%(t-1) + (t-1)) % (t-1);
        }
        else{
            vec[i] = ((tp + pp)%(t+1) + (t+1)) % (t+1);
        }
        pp *= 2;
        pm *= 2;
        pp %= (t+1);
        pm %= (t-1);
    }
    reverse(ALL(vec));
    For(i,0,n){
        while(vec[i]){
            vec[i] %= __builtin_popcount(vec[i]);
            ans[i]++;
        }
        cout << ans[i] << endl;
    }
}
int main(){
    Main();
    /*
    東方風神録は神が出てくるので当然神ゲー
    */
    return 0;
}