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 lol;

//マクロ
//forループ関係
//引数は、(ループ内変数,動く範囲)か(ループ内変数,始めの数,終わりの数)、のどちらか
//Dがついてないものはループ変数は1ずつインクリメントされ、Dがついてるものはループ変数は1ずつデクリメントされる
#define REP(i,n) for(lol i=0;i<lol(n);i++)
#define REPD(i,n) for(lol i=n-1;i>=0;i--)
#define FOR(i,a,b) for(lol i=a;i<=lol(b);i++)
#define FORD(i,a,b) for(lol i=a;i>=lol(b);i--)
//xにはvectorなどのコンテナ
#define ALL(x) x.begin(),x.end() //sortなどの引数を省略したい
#define SIZE(x) lol(x.size()) //sizeをsize_tからllに直しておく
//定数
#define INF 1000000000000 //10^12:極めて大きい値,∞
#define MOD 1000000007 //10^9+7:合同式の法
#define MAXR 100000 //10^5:配列の最大のrange(素数列挙などで使用)
//最大値最小値
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
int gcd(int a,int b){return b?gcd(b,a%b):a;}

int popcountn(int);
lol rec(lol, int);

int main()
{
    lol result{0};
    lol n; cin >> n;
    //int k; cin >> k;
    bitset<32> a;
    cin >> a;
    //a.flip(0);
    //result = popcountn(n);
    //cout << a << endl;

    FORD(i, n-1, 0)
    {
        int k{0};
        a.flip(i);
        result = rec(a.to_ullong(), k);
        cout << result << endl;
        a.flip(i);
    }

    
    return 0;
}

int popcountn(int n)
{
    return bitset<64>(n).count();
}


lol rec(lol n, int k)
{
    if(n == 0)
        return k;
    else
    {
        int pn = popcountn(n);
        k++;
        return rec(n%pn, k);
    }
    
}