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)

#include <bits/stdc++.h>
using namespace std;

//定数
const long long MOD1=1000000007;
const long long MOD2=998244353;
const long double PI=3.1415926535897932;
const long long MAXLL=9223372036854775807;
const long long INF=2305843009213693951;
const long long dx[]={0,1,0,-1,1,-1,1,-1};
const long long dy[]={1,0,-1,0,1,1,-1,-1};

//型名省略
#define ll long long
#define ull unsigned long long
#define ld long double
#define uld unsigned long double
#define pll pair<long long,long long>
#define tll(n) array<long long,(int)(n)>
#define pcc pair<char,char>
#define vl vector<long long>
#define vvl vector<vector<long long>>
#define vc vector<char>
#define vvc vector<vector<char>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define vp vector<pair<long long,long long>>
#define vt(n) vector<array<long long,(int)(n)>>
#define umap unordered_map
#define uset unordered_set
#define Lqueue priority_queue<long long>
#define Squeue priority_queue<long long,vector<long long>,greater<long long>>
#define Fi first
#define Se second
#define mp make_pair
#define eb emplace_back

//マクロ
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define rbf(a,x) for(auto& a:x)
#define rep(i,n) for(long long i=0;i<(long long)(n);i++)
#define rep2(i,s,n) for(long long i=(s);i<(long long)(n);i++)
#define bitrep(i,s,n) for(long long i=(s);i<(1LL<<(n));i++)
#define Maxe(x) *max_element((x).begin(),(x).end())
#define Mine(x) *min_element((x).begin(),(x).end())
#define Size(x) ((long long)(x).size())
#define lin(s) getline(cin,(s))
#define Ssort(v) sort((v).begin(),(v).end(),[](auto &left,auto &right){return left.second<right.second;})

//Yes,No
void Yes(bool a){cout<<(a?"Yes":"No")<<endl;}
//YES.NO
void YES(bool a){cout<<(a?"YES":"NO")<<endl;}

//MAX,MIN
template<class T,class U> auto max(T a,U b){return a>b?a:b;}
template<class T,class U> auto min(T a,U b){return a<b?a:b;}

//最大公約数,最小公倍数
long long gcd(long long a,long long b){return b?gcd(b,a%b):a;}
long long lcm(long long a,long long b){return a/gcd(a,b)*b;}

//累乗
template<typename t>
constexpr t mypow(t a,long long b){
    if(b==0)return 1;
    if(a==0)return 0;
    t x=1;
    while(b>0){
        if(b&1LL)x*=a;
        a*=a;
        b>>=1LL;
        }
    return x;
}
#define pow mypow

//chmin,chmax
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return true;}return false;}
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return true;}return false;}

//組み合わせ O(r)
template<typename t>
constexpr t nCr(t n,long long r){
    if(r==0)return 1;
    if(n==0)return 0;
    if(n<r)return 0;
    t x=1;
    for(long long i=1;i<=r;i++){
        x*=n-i+1;
        x/=i;
    }
    return x;
}

ll solve(ll a){
    ll c=__builtin_popcount(a);
    if(c==0)return 1;
    return solve(a%c)+1;
}

//main関数
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll n;
    string x;
    cin>>n>>x;
    ll cnt=0;
    rep(i,n)if(x[i]=='1')cnt++;
    vl d1(n),d2(n);
    d1[0]=d2[0]=1;
    rep2(i,1,n){
        d1[i]=d1[i-1]*2%(cnt+1);
        if(cnt-1!=0)d2[i]=d2[i-1]*2%(cnt-1);
    }
    ll sum1=0;
    ll sum2=0;
    rep(i,n){
        if(x[n-1-i]=='1'){
            sum1+=d1[i];
            sum1%=cnt+1;
            if(cnt-1!=0){
                sum2+=d2[i];
                sum2%=cnt-1;
            }
        }
    }
    rep(i,n){
        ll res;
        if(x[i]=='1'){
            res=sum2;
            res-=d2[n-1-i];
            res+=cnt-1;
            if(cnt-1!=0)res%=cnt-1;
            else{
                cout<<0<<endl;
                continue;
            }
        }else{
            res=sum1;
            res+=d1[n-1-i];
            res%=cnt+1;
        }
        cout<<solve(res)<<endl;
    }
}