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>
#define rep(i,a,b) for (i=a;i<b;i++)
#define rep2(i,a,b) for (i=a;i>=b;i--)
// #define mod 1000000007
#define mod2 998244353
#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define INF  1e9+5
#define f first
#define s second
#define endl '\n'
#define ll long long
#define ii pair <int,int>
#define pll pair <ll,ll>
#define vi vector <int>
#define vl vector <ll>
#define vvi vector < vi >
#define vii vector < ii >
#define vpl vector < pll >
#define vvii vector < vii >
#define vll vector < ll >
#define vb  vector <bool>
#define pb push_back
#define mk make_pair
#define eb emplace_back
#define pg priority_queue<ll, vector<ll>, greater<ll>>
#define pq priority_queue<ll>
#define sz(a) a.size()
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define cset(a) __builtin_popcountll(a)
#define FILL(a, b, T) fill_n((T *)a, sizeof(a) / sizeof(T), b)
// #define debug(x) cerr << #x << " is " << x << endl;
using namespace std;
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
#define pis pair<ll,pll>
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << endl;
    err(++it, args...);
}
const auto start_time = std::chrono::high_resolution_clock::now();
void timecalculater(){
    #ifndef ONLINE_JUDGE
    auto end_time = std::chrono::high_resolution_clock::now();
    std::chrono::duration<double> diff = end_time-start_time;
    cerr<<"Time Taken : "<<diff.count()<<" s\n";
    #endif
}
template <typename A, typename B>
string to_string(pair<A, B> p);
 
string to_string(const string& s) {
        return '"' + s + '"';
}
 
string to_string(const char* s) {
        return to_string((string) s);
}
 
string to_string(bool b) {
        return (b ? "true" : "false");
}
 
string to_string(vector<bool> v) {
        bool first = true;
        string res = "{";
        for (int i = 0; i < static_cast<int>(v.size()); i++) {
                if (!first) {
                        res += ", ";
                }
                first = false;
                res += to_string(v[i]); 
        }
        res += "}";
        return res;
}
 
template <size_t N>
string to_string(bitset<N> v) {
        string res = "";
        for (size_t i = 0; i < N; i++) {
                res += static_cast<char>('0' + v[i]);
        }
        return res;
}
 
template <typename A>
string to_string(A v) {
        bool first = true;
        string res = "{";
        for (const auto &x : v) {
                if (!first) {
                        res += ", ";
                }
                first = false;
                res += to_string(x);
        }
        res += "}";
        return res;
}
 
template <typename A, typename B>
string to_string(pair<A, B> p) {
        return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
 
void debug_out() { 
        cerr << endl; 
}
 
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
        cerr << " " << to_string(H);
        debug_out(T...);
}
 
#ifndef ONLINE_JUDGE
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...)
#endif
struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }
 
    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};
// unordered_map<long long, int, custom_hash> safe_map;
// FOR VIEWERS, CODE STARTS HERE
static ll poww(ll a,ll b, ll mod)
{
    ll x=1,y=a;
    while(b>0)
    {   
        if(b%2)
        {
            x = (x* y)%mod;
        }
        y = (y*y)%mod;
        b/=2;
    }
    return (x%mod);
}
ll add(ll x,ll y, ll mod)
{
    return (x+y+mod)%mod;
}
int main()
{
    FIO;
    ll t, q, n, m, prev, maxx,  i, j, k1, k2, k3, k, x, y, flag, sum, cnt, ans, minn, cur;
    cin>>n;
    string ss;
    cin>>ss;
    reverse(all(ss));
    ans = 0;
    ll ans2 = 0;
    cnt = 0;
    for(auto p: ss)
        if( p == '1')
            cnt++;
    n = cnt;
    rep(i,0, sz(ss))
    {
        if( ss[i] == '1')
        {
            ans = add(ans, poww(2, i, n - 1), n - 1);
            ans2 = add(ans2, poww(2, i, n + 1), n + 1);
        }
    }
    vll res;
    rep(i,0,sz(ss))
    {
        if( ss[i] == '1') // remove then ans1
        {
            
            cur = add(ans, -poww(2, i, n - 1), n - 1);
        }
        else
        {
            cur = add(ans2, poww(2, i, n + 1), n + 1);
        }
        cnt = 1;

        while( cur > 0)
        {
            cur %= cset(cur);
            cnt++;
        }
        res.pb(cnt);
    }
    reverse(all(res));
    for(auto p: res)
        cout<<p<<endl;
    return 0;    
}