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>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;

#define INX(x) (cin >> x)
#define OUTX(x) (cout << x << endl)

#define ADD emplace_back
#define MP make_pair
#define VVEC(type) vector<vector<type>>


map<ll, ll> pcnt;
ll popcount(ll x)
{
    if(pcnt.find(x) == pcnt.end())
    {
        ll result = 0;
        ll xcopy = x;
        while(x)
        {
            if(x & 1)
            {
                result++;
            }
            x >>= 1;
        }
        pcnt.emplace(MP(xcopy, result));
        return result;
    }
    else
    {
        return pcnt[x];
    }
}

map<ll, ll> fcnt;
ll f(ll x)
{
    if(x == 0)
    {
        return 0;
    }
    else if(fcnt.find(x) == fcnt.end())
    {
        ll p = popcount(x);
        ll result = f(x % p) + 1;
        fcnt.emplace(MP(x, result));
        return result;
    }
    else
    {
        return fcnt[x];
    }    
}


int main()
{
    ll N;
    INX(N);

    string Xstring;
    INX(Xstring);
    ll X = stoll(Xstring, 0, 2);

    for(ll i = (1 << (N - 1)); i != 0; i >>= 1)
    {
        ll result = f(X ^ i);
        OUTX(result);
    }


    return 0;
}