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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int ll
typedef vector<int> vi;
// #define endl '\n'
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define rf(i,a,b) for(int i=a;i>=b;i--) 
#define fast_io ios_base::sync_with_stdio(0);
#define show(x) cout<<x<<endl;
#define pret(x) cout<<x<<endl; return 0;
#define disp(x) cout<<x<<" ";
#define let(x,y) cout<<x<<" "<<y<<endl;
// ll MOD=1e9+7;
const int maxn=2e5+5;

int ans[maxn];
int val[2];

ll power(ll a, ll b, ll MOD){
	if(MOD == 0) return 0;
    ll x = 1, y = a;
    while (b > 0){
        if (b%2){
            x = (x*y)%MOD;
        }
        y = (y*y)%MOD;
        b /= 2;
    }
    return x%MOD;
}  

int calc(int x){
	int temp = __builtin_popcount(x);
	temp = x%temp;
	return 1+ans[temp];
}

int32_t main(){
    fast_io;
    int n;
    cin>>n;
    string s;
    cin>>s;
    fo(i,1,maxn-1){
    	ans[i] = calc(i);
    }
    // show(ans[8]);
    vi v;
    int tot = 0;
    for (int i = 0; i < n; ++i)
    {
    	v.emplace_back(s[i]-'0');
    	tot += v[i];
    }
    if(tot == 1){

    }
    // show(tot);
    for (int i = 0; i < n; ++i)
    {
    	if(v[i]) {
    		val[0] += power(2,n-1-i,tot-1);
    		val[1] += power(2,n-1-i,tot+1);
    	}
    }
    // let(val[0],val[1]);
    for (int i = 0; i < n; ++i)
    {
    	if(v[i]){
    		if(tot == 1) {
    			show(0);
    			continue;
    		}
    		int temp = val[0] - power(2,n-1-i,tot-1);
    		// let(i,temp);
    		show(1 + ans[temp]);
    	}
    	else{
    		int temp = val[1] + power(2,n-1-i,tot+1);
    		temp %= (tot+1);
    		// let(i,temp);
    		show(1+ ans[temp]);
    	}
    }
    return 0;
}