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;

typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<vll> vvll;

#define all(x) (x).begin(), (x).end()
#define MOD 1000000007
#define forn(i, n) for (int i = 0; i < int(n); ++i)
#define fore(i, l, r) for (int i = int(l); i < int(r); ++i)
 
template<class A, class B> ostream& operator << (ostream& out, const pair<A, B> &a) {
    return out <<  "(" << a.first << ", " << a.second << ")";
}
 
template<class A> ostream& operator << (ostream& out, const vector<A> &a) {			
    out << "[";
    for (auto it = a.begin(); it != a.end(); ++it) {
        if (it != a.begin())
            out << ", ";
        out << *it;
    }
    return out << "]";
}
int arr[200001];

void solve(){
    ll n;
    cin >> n;
    string x;
    cin >> x;

    //n = 1
    if(n==1){
        if(x=="0") cout<<1<<"\n";
        else cout<<0<<"\n";
        return;
    }
    arr[0] = 0;

    for(int i=1;i<=200000;i++){
        arr[i] = arr[i%__builtin_popcount(i)]+1; 
    }
    ll curr1 = 0;
    ll curr2 = 0;
    vll pw1={1};
    vll pw2={1}; 
    ll cnt = 0;
    for(auto cc : x){
        if(cc == '1') cnt++;
    }
    if(cnt == 0){
        for(int i=0;i<n;i++)
            cout<<1<<"\n";
        return; 
    }
    for(int i = n-1;i >= 0 ;i--){
        if(x[i] == '1'){
            curr1+=pw1.back();
            curr2+=pw2.back();
            curr2%=cnt+1;
            curr1%=cnt-1;
        }
        if(cnt!=1)
        pw1.push_back((2LL*pw1.back())%(cnt-1));
        pw2.push_back((2LL*pw2.back())%(cnt+1));
        // pw1.push_back((2LL*pw1.back()));
        //pw2.push_back((2LL*pw2.back()));
    } 

    reverse(all(pw1));
    reverse(all(pw2));
    for(int i=0;i<n;i++){ 
        if(x[i] == '0'){
            cout<<arr[(curr2 + pw2[i+1])%(cnt+1)]+1<<"\n";
        }
        else{
            if(cnt == 1)
            cout<<0<<"\n";
            else
            cout<<arr[(curr1 - pw1[i+1] + (cnt-1))%(cnt-1)]+1<<"\n";
        }
    }

}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll t=1;
   // cin >> t;
    while(t--){
        solve();
    }

	return 0;
}