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>
#include <random>
#include <ctime>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//#pragma GCC optimize("-O3")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("avx2")
#define ll long long
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define pll pair<ll,ll>
#define ld double
#define pld pair<ld,ld>
#define pii pair<int,int>
#define sqr(a) ((a)*(a))
#define all(v) v.begin(),v.end()

using namespace std;
//using namespace __gnu_pbds;

ll MOD = 998244353;
const ll INF = 1e18;
const ld E = 1e-18;
const ld PI=acos(-1);
mt19937_64 rnd(time(0));
ll p1 = 31, p2 = 37, p3 = 7, p4 = rnd() % 100 + 1, md1 = MOD, md2 = 998244353, md3 = 998244357, md4 = rnd() % 1000000000;

//using ord_set=tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;

ll n, m, T, a[1000001];

ll binpow(ll h, ll r, ll md = MOD)
{
    ll l = 1;
    while (r)
    {
        if (r & 1) l *= h, l%=md;
        h *= h;
        h %= md;
        r /= 2;
    }
    return l;
}

ll gcd(ll x, ll y)
{
    if (x < y) swap(x, y);
    while (x && y)
    {
        x %= y;
        swap(x, y);
    }
    return x + y;
}

ll check(int h)
{
    ll j=0;
    while (h)
    {
        ll x=h, kl=0;
        while (x)
        {
            kl+=x&1;
            x/=2;
        }
        h%=kl;
        j++;
    }
    return j;
}

int main()
{
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    //freopen("capitals.in","r",stdin);
    //freopen("capitals.out","w",stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    string s;
    cin >> s;
    s=' '+s;
    ll kl=0;
    for (int i = 1; i <= n; i++)
    {
        kl+=s[i]-48;
    }
    ll k1=0, k2=0;
    for (int i = n, j = 0; i > 0; i--, j++)
    {
        if (s[i]=='1')
        {
            if (kl>1) k1=(k1+binpow(2,j,kl-1))%(kl-1);
            k2=(k2+binpow(2,j,kl+1))%(kl+1);
        }
    }
    for (int i = 1, j = n-1; i <= n; i++, j--)
    {
        if (s[i]=='1')
        {
            if (kl==1)
            {
                cout << "0\n";
                continue;
            }
            ll p=k1;
            //cout << binpow(2,j,kl-1) << " " << kl-1 << "\n";
            p-=binpow(2,j,kl-1);
            if (p<0) p+=kl-1;
            cout << 1+check(p) << "\n";
        }else
        {
            ll p=k2;
            p=(p+binpow(2,j,kl+1))%(kl+1);
            cout << 1+check(p) << "\n";
        }
    }
    return 0;
}

/*
3
4 2
1 1
1 0
3 2
2 2

3 1
1 1
1 0
3 2

3 2
1 1
1 0
3 2
*/