#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>   
using namespace std;
using namespace __gnu_pbds;
typedef long long ll; 
#define int long long
typedef unsigned long long lu;
typedef vector<ll> v;
typedef vector<vector<ll> > vv;
typedef vector<string>  vs;
typedef vector<pair<ll,ll>> vpr;
typedef vector<bool>vb;
typedef vector<double>vd;
typedef long double ld;
#define f(i,n) for(ll i = 0; i < n; i++)
#define ff(i,n) for(ll i=1;i<=n;i++)
#define pb push_back
#define mp make_pair
#define endl "\n"
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define amax(x,y) if(y>x) x=y;
#define amin(x,y) if(y<x)x=y;
#define bg(x) x.begin()
#define sz(x) (ll)x.size()
#define in(x,n) for(ll i=0;i<n;i++)cin>>x[i]
#define out(x,n) for(ll i=0;i<n;i++)cout<<x[i]<<" " 
#define mxt(a) *(max_element(a.begin(),a.end()))
#define mnt(a) *(min_element(a.begin(),a.end())
#define tc ll t;cin>>t;while(t--)
typedef pair<ll,ll> pi;
#define yes cout<<"YES\n";
#define no cout<<"NO\n";
#define yesno(f) if(f) yes else no
const v dx = {1, -1, 0, 0};
const v dy = {0, 0, 1, -1};
const ld PI = 2 * acos(0.0);
ll cel(ll x1,ll y1){if((x1%y1)==0)return x1/y1;else return x1/y1+1;}
ll power(ll a,ll b,ll m)
{
    if(b==0)
        return 1;
    ll d=power(a,b/2,m);
    d=(d*d)%m;
    if(b&1)
        d=(d*a)%m;
    return d;
}
const ll mod=1e9+7;
int MOD(int a)
{
    if(a<0)
        a+=mod;
    if(a>=mod)
        a%=mod;
    return a;
}
// set_name.find_by_order(k) It returns to an iterator to the kth element (counting from zero) in the set in O(logn) time
// set_name.order_of_key(k) It returns to the number of items that are strictly smaller than our item k in O(logn) time.
/*string operations :
str.substr (x,y) : returns a substring str[x],str[x+1],...str[x+y-1]
str.substr (x) : returns a substring str[x],... end of string
str.find(qtr) : returns the first occurenece of qtr in str */
int fun(int i)
{
    int c=0;
    while(i)
    {
        c+=(i%2);
        i/=2;
    }
    return c;
}
int32_t main()
{
    ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    int n;cin>>n;
    string s;cin>>s;
    int c=0;
    for(int i=0;i<n;i++)
    {
        if(s[i]=='1')
            c++;
    }
    v ans(n+1,0);
    for(int i=1;i<=n;i++)
    {
        ans[i]=ans[i%fun(i)]+1;
    }
    // out(ans,n);
    vv powers(n,v(2,1)),pre(n,v(2,0)),suf(n,v(2,0));
    int c1=c+1,c2=max(c-1,1ll);
    bool flg=false;
    if(c==1)
        flg=true;
    for(int i=1;i<n;i++)
    {
        powers[i][0]=(2*powers[i-1][0])%c1;
        powers[i][1]=(2*powers[i-1][1])%c2;
    }
    // v pre(n),suf(n);
    if(s[0]=='1')
    {
        pre[0][0]=powers[n-1][0];
        pre[0][1]=powers[n-1][1]; 
    }
    if(s[n-1]=='1')
    {
        suf[n-1][0]=powers[0][0];
        suf[n-1][1]=powers[0][1];
    }
    for(int i=1;i<n;i++)
    {
        if(s[i]=='0')
        {
            pre[i][0]=pre[i-1][0];
            pre[i][1]=pre[i-1][1];
            continue;
        }
        else
        {
            pre[i][0]=(pre[i-1][0]+powers[n-1-i][0])%c1;
            pre[i][1]=(pre[i-1][1]+powers[n-1-i][1])%c2;
        }
    }
    for(int i=n-2;i>=0;i--)
    {
        if(s[i]=='0')
        {
            suf[i][0]=suf[i+1][0];
            suf[i][1]=suf[i+1][1];
            continue;
        }
        else
        {
            suf[i][0]=(suf[i+1][0]+powers[n-i-1][0])%c1;
            suf[i][1]=(suf[i+1][1]+powers[n-i-1][1])%c2;
        }
    }
    for(int i=0;i<n;i++)
    {

              int ch=0;
        if(s[i]=='1')
        {
            if(flg)
            {
                cout<<0<<endl;
                continue;
            }
            // int ch=0;
            if(i!=0)
                ch+=pre[i-1][1];
            if(i!=(n-1))
                ch+=suf[i+1][1];
            ch%=c2;
        }
        else
        {
                if(i!=0)
                    ch+=pre[i-1][0];
                if(i!=(n-1))
                    ch+=suf[i+1][0];
                ch%=c1;
            ch+=powers[n-i-1][0];
            ch%=c1;

        }
        
        // cout<<ch<<endl;
        cout<<ans[ch]+1<<endl;
    }
    return 0;
}

Measured memory_bytes: 37928