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 <bits/stdc++.h>
#define rep(i,n) for(ll i=0; i<(n); i++)
#define chmin(a,b) a = min(a,b)
#define chmax(a,b) a = max(a,b)
using namespace std;
using ll = long long;
using vi = vector<int>;
using vv = vector<vi>;
using P  = pair<int,int>;

const ll INF = 1001001001;
const int MOD = 1000000007;

int main(){
    int n;
    string x;
    cin>>n>>x;
    
    
    vi used(n+10,-1);
    
    int popx = 0;
    rep(i,n)if(x[i]=='1')popx++;
    
    if(popx==0){
        rep(i,n)cout<<1<<endl;
        return 0;
    }
    
    vi popn(2);
    popn[0] = popx+1;
    popn[1] = popx-1;
    
    //前もって2^kのmod popx \pm 1を計算しておく
    
    vv modpop(2, vi (n));
    
    modpop[0][n-1] = 1%popn[0];
    if(popn[1]!=0)modpop[1][n-1] = 1%popn[1];
    
    for(int i=n-2; i>=0; i--){
        modpop[0][i] = 2*modpop[0][i+1]%popn[0];
        if(popn[1]!=0)modpop[1][i] = 2*modpop[1][i+1]%popn[1];
    }
    
    for(int j=0; j<n; j++){
        int ans = 1;
        int count = 0;
        int c = 0;
        if(x[j]=='1')c = 1;
        
        if(popn[1]==0 && c==1){
            cout<<0<<endl;
            continue;
        }
        
        rep(i,n){
            if(i==j && x[i]=='0'){
                count += modpop[0][i];
                
            }
            else if(i!=j && x[i]=='1'){
                count += modpop[c][i];
                
            }
            
            count %= popn[c];
        }
        
        
        if(count==0){
            cout<<1<<endl;
            continue;
        }
        
        int num = count;
        
        if(used[count]!=-1){
            cout << ans + used[count] << endl;
            continue;
        }
        
        while(num){
            
            if(used[num]!=-1){
                ans += used[num];
                break;
            }
            //std::cout << std::bitset<20>(num)<<endl; // 00001001
            //cout << num<<endl;
            int p = 0;
            int b = num;
            
            while(b){
                p += (b&1);
                b /= 2;
            }
            
            num %= p;
            ans++;
        }
        
        if(used[count] == -1)used[count]=ans-1;
        
        cout << ans << endl;
    }
    
    
    
    return 0;
}