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)

/**
 *    author:  otera    
**/
#include<iostream>
#include<string> 
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<algorithm> 
#include<functional>
#include<iomanip>
#include<queue>
#include<deque>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>

// #include <boost/multiprecision/cpp_dec_float.hpp>
// #include <boost/multiprecision/cpp_int.hpp>
// #include <boost/rational.hpp>

// namespace mp = boost::multiprecision;
// // 任意長整数型
// using Bint = mp::cpp_int;

using namespace std;

#define int long long
typedef long long ll;
typedef unsigned long long ul;
typedef unsigned int ui;
typedef long double ld;
const int inf=1e9+7;
const ll INF=1LL<<60 ;
const ll mod=1e9+7 ;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef complex<ld> Point;
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<int, int> P;
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;
#define fr first
#define sc second
#define all(c) c.begin(),c.end()
#define pb push_back
#define debug(x)  cerr << #x << " = " << (x) << endl;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int f(int a) {
    int co = __builtin_popcount(a);
    return a % co;
}

void solve() {
	int n; string x; cin >> n >> x;
    int cnt = 0;
    rep(i, n) {
        if(x[i] == '1') ++ cnt;
    }
    static int p[3][200200];
    rep(i, 3) p[i][0] = 1;
    p[0][0] %= (cnt + 1);
    for(int i = 1; i <= 200000; ++ i) {
        p[0][i] = p[0][i - 1] * 2 % (cnt + 1);
    }
    if(cnt > 0) {
        p[1][0] %= cnt;
        for(int i = 1; i <= 200000; ++ i) {
            p[1][i] = p[1][i - 1] * 2 % cnt;
        }
    }
    if(cnt - 1 > 0) {
        p[2][0] %= (cnt - 1);
        for(int i = 1; i <= 200000; ++ i) {
            p[2][i] = p[2][i - 1] * 2 % (cnt - 1);
        }
    }
    //cnt + 1
    int res0 = 0;
    per(i, n) {
        res0 += (x[i] == '1' ? p[0][n - 1 - i] : 0);
        res0 %= (cnt + 1);
    }
    //cnt - 1
    int res2 = 0;
    if(cnt - 1 > 0) {
        per(i, n) {
            res2 += (x[i] == '1' ? p[2][n - 1 - i] : 0);
            res2 %= (cnt - 1);
        }
    }
    rep(i, n) {
        int co = cnt;
        if(co == 0) {
            cout << 0 << endl;
            return;
        }
        if(x[i] == '0') {
            //0
            co ++;
            int cur = (res0 + p[0][n - 1 - i]) % (cnt + 1);
            int ans = 1;
            while(cur != 0) {
                ++ ans;
                cur = f(cur);
            }
            cout << ans << endl;
        } else {
            //2
            co --;
            if(co == 0) {
                cout << 0 << endl;
                return;
            }
            int cur = (res0 - p[2][n - 1 - i]) % (cnt - 1);
            int ans = 1;
            while(cur != 0) {
                ++ ans;
                cur = f(cur);
            }
            cout << ans << endl;
        }
    }
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(10);
	//int t; cin >> t; rep(i, t)solve();
	solve();
    return 0;
}