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 <bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,n) for(ll i = 0; i < (ll) n; i++)

int ppc(ll n){
    ll ans = 0;
    while(n>0){
        if(n%2==1){
            ans++;
        }
        n=n/2;
    }
    return ans;
}
int f(ll n) {
    int ans=0;
    while(n>0){
        n=n%ppc(n);
        ans++;
    }
    return ans;
}

int main() {
    ll N;
    string X;
    cin>>N>>X;

    ll X10=0;
    ll base=1;
    ll x=atoi(X.c_str());
    while(x>0){
        X10+=(x%10)*base;
        x/=10;
        base*=2;
    }

    vector<ll> a(N);
    rep(i,N){
        if(X[i]=='0'){
            a[i]=X10+pow(2,N-1-i);
        }
        else if(X[i]=='1'){
            a[i]=X10-pow(2,N-1-i);
        }
        cout<<f(a[i])<<endl;
    }
}