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: Accepted)

/*** author: yuji9511 ***/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using lpair = pair<ll, ll>;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define rep(i,m,n) for(ll i=(m);i<(n);i++)
#define rrep(i,m,n) for(ll i=(m);i>=(n);i--)
#define printa(x,n) for(ll i=0;i<n;i++){cout<<(x[i])<<" \n"[i==n-1];};
void print() {}
template <class H,class... T>
void print(H&& h, T&&... t){cout<<h<<" \n"[sizeof...(t)==0];print(forward<T>(t)...);}
ll power(ll x, ll n, ll M){
    if(n == 0) return 1LL;
    ll res = power(x * x % M, n/2, M);
    if(n % 2 == 1) res = res * x % M;
    return res;
}

void solve(){
    ll N;
    cin >> N;
    string X;
    cin >> X;
    ll c1 = 0;
    rep(i,0,N){
        if(X[i] == '1') c1++;
    }
    ll amari1 = 0, amari2 = 0;
    ll pow_val = 1;
    rrep(i,N-1,0){
        amari1 += pow_val * (X[i] - '0') % (c1+1);
        amari1 %= (c1+1);
        pow_val *= 2;
        pow_val %= (c1+1);
    }
    pow_val = 1;
    rrep(i,N-1,0){
        if(c1 <= 1) break;
        amari2 += pow_val * (X[i] - '0') % (c1-1);
        amari2 %= (c1-1);
        pow_val *= 2;
        pow_val %= (c1-1);
    }
    rep(i,0,N){
        if(X[i] == '1'){
            if(c1 == 1){
                print(0);
                continue;
            }
            ll pow_val = power(2, N-1-i, c1-1);
            ll cur = (amari2 - pow_val + (c1-1)) % (c1-1);
            ll ans = 1;
            while(cur > 0){
                ll cnt = __builtin_popcount(cur);
                cur %= cnt;
                ans++;
            }
            print(ans);
        }else{
            ll pow_val = power(2, N-1-i, c1+1);
            ll cur = (amari1 + pow_val) % (c1+1);
            ll ans = 1;
            while(cur > 0){
                ll cnt = __builtin_popcount(cur);
                cur %= cnt;
                ans++;
            }
            print(ans);
        }
    }



    

}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
}