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: Wrong Answer)

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

using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))
#define FOR(i, begin, end) for (int i = (begin), i##_end_ = (end); i < i##_end_; i++)
#define IFOR(i, begin, end) for (int i = (end)-1, i##_begin_ = (begin); i >= i##_begin_; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
#ifdef LOCAL
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define eprintf(...) 42
#endif

int main()
{
    lint n;
    cin >> n;
    string x;
    cin >> x;
    lint cnt = 0;
    REP(i,n){
        cnt += (x[i] == '1');
    }
    int v = (x[n - 1] == '1');
    lint ans, tmp;
    vector<lint> l(n+1),r(n+1);
    lint ll = cnt - 1, rr = cnt + 1;
    ll += (ll == 0);
    rr += (rr == 0);
    l[1] = v % (ll);
    r[1] = v % (rr);
    vector<lint> addl(n + 1), addr(n + 1);
    addl[1] = 1%ll;
    addr[1] = 1%rr;
    FOR(i,1,n) {
        addl[i + 1] = (addl[i] * 2) % ll;
        addr[i + 1] = (addr[i] * 2) % rr;
        if(x[n-i-1]=='0'){
            l[i + 1] = l[i];
            r[i + 1] = r[i];
        }else{
            l[i + 1] = (l[i] +addl[i+1]) % ll;
            r[i + 1] = (r[i] +addr[i+1]) % rr;
        }
    }
    /*
    FOR(i,1,n+1){
        cout << l[i] << " ";
    }
    cout << "\n";
    FOR(i, 1, n + 1)
    {
        cout << r[i] << " ";
    }
    cout << "\n";
    */
    vector<lint> f(n + 1),g(n+1);
    FOR(i,1,n+1){
        lint k = i;
        lint bi = __builtin_popcountl(k);
        //cout << bi << " ";
        lint tmp = (k % bi);
        if(tmp==0){
            g[i] = 1;
        }else{
            g[i] = g[tmp] + 1;
        }
    }
    //cout << "aaa";
    /*
    REP(i,n+1){
        cout << g[i] << " ";
    }
    cout  << "\n";
    */
    IFOR(i,1,n+1){
        lint z;
        if(x[n-i]=='0'){
            z = (r[n-1]+rr+addr[i])%rr;
        }else{
            z = (l[n - 1] + ll - addl[i]) % ll;
        }
        //cout << z << "\n";
        cout << g[z]+1 << "\n";
    }
    return 0;
}