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>
#define rep(i,n) for (int i = 0; i < (int)(n); i++)
#define REP(i,n) for (int i = 1; i < (int)(n); i++)
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define debug(var) do{cout << #var << " : "; view(var);}while(0)
template<class T> bool chmin(T &a, T b) {if(a>b) {a=b;return 1;}return 0;}
template<class T> bool chmax(T &a, T b) {if(a<b) {a=b;return 1;}return 0;}
using namespace std;
template<class T> void view(T e) {cout << e << endl;}
template<class T> void view(const vector<T> &v) {for(const auto &e : v){cout << e << " ";} cout << endl;}
template<class T> void view(const vector<vector<T>> &vv) {for(const auto &v : vv){view(v);}}
using vint = vector<int>;
using vvint = vector<vector<int>>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<int,int>;
const int inf = 1e9;
const ll inf_l = 1e18;
const int MAX = 1e5;

long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

int f(string s, int div) {
    int n = s.size();
    ll x = 0;
    rep(i,n) {
        if (s[i] == '1') {
            x += modpow(2,n-1-i,div);
            x %= div;
        }
    }
    return x%div;
}

int f_(int x, int ct = 1) {
    if (x == 0) return ct;
    int tmp = x % __builtin_popcount(x);
    if (tmp == 0) return ct+1;
    return f_(tmp,ct+1);
}

int main() {
    int n; cin >> n;
    string s; cin >> s;
    int div = 0;
    rep(i,n) if (s[i] == '1') div++;
    int x_plus = f(s,div+1);
    int x_minus = f(s,div-1);
    rep(i,n) {
        int div_tmp = div;
        string s_tmp = s;
        if (s[i] == '0') {
            s_tmp[i] = '1';
            ll x = x_plus+modpow(2,n-1-i,div+1);
            x %= div+1;
            // debug(x);
            cout << f_(x) << endl;
        }
        if (s[i] == '1') {
            s_tmp[i] = '0';
            if (div-1 == 0) cout << 0 << endl; 
            else {
                ll x = x_minus-modpow(2,n-1-i,div-1);
                if (x < 0) x += div-1;
                x %= div-1;
                // debug(x);
                cout << f_(x) << endl;
            }
        }
        // debug(s_tmp);
    }
}