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<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>
#include<complex>
#include <cassert>

using namespace std;

#define ll long long
#define ld long double
#define INF 1e9
#define LINF (ll)INF*INF
#define MOD 1000000007
#define rep(i,n) for(int i=0;i<(n);i++)
#define loop(i,a,n) for(int i=a;i<(n);i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)

#define int ll //!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

const double PI=acos(-1);
const double EPS=1e-9; //うまくいかなかったらゆるめる

typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int,int> pii;
typedef pair<pii,int> ppi;
typedef pair<int,pii> pip;
typedef vector<pii> vp;
typedef vector<vi> vvi;

int gcd(int a, int b){if(b==0) return a;return gcd(b,a%b);}
int lcm(int a, int b){return a/gcd(a,b)*b;}

map<int,int>ma;

int f(int n){
    if(n == 0)return 0;
    if(ma[n])return ma[n];
    int cnt = 0;
    rep(i,60)if(n & (1LL<<i))cnt++;
    return ma[n] = f(n%cnt) + 1;
}

signed main(void) {
    int n;
    cin >> n;
    string s;
    cin >> s;
    ma[1] = 1;
    ma[2] = 1;
    int cnt = 0;
    rep(i,n)if(s[i] == '1')cnt++;
    vi ans(n);
    int a = cnt-1;
    int b = cnt+1;
    int suma = 0;
    int sumb = 0;
    if(a)rep(i,n){
        suma *= 2;
        suma += s[i]-'0';
        suma %= a;
    }
    if(a)rep(i,n){
        sumb *= 2;
        sumb += s[i]-'0';
        sumb %= b;
    }
    rep(i,n){
        int tmp = 1;
        if(s[i] == '1'){//a
            if(a == 0)ans[i] = 1;
            else{
                rep(_,n-1-i)tmp *= 2, tmp %= a;
                int t = suma + a - tmp;
                ans[i] = f(t%a)+1;
            }
        }else{//b
            if(b == 0)ans[i] = 1;
            else{
                rep(_,n-1-i)tmp *= 2, tmp %= b;
                int t = sumb + tmp;
                ans[i] = f(t%b)+1;
            }
        }
    }
    rep(i,n)cout << ans[i] << endl;
}