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;
#define ll long long
#define FOR(i,l,r) for(ll i=(l);i<(r);++i)
#define REP(i,n) FOR(i,0,n)
#define REPS(i,n) FOR(i,1,n+1)
#define RFOR(i,l,r) for(ll i=(l);i>=(r);--i)
#define RREP(i,n) RFOR(i,n-1,0)
#define RREPS(i,n) RFOR(i,n,1)
#define pb push_back
#define eb emplace_back
#define popcnt __builtin_popcount
#define SZ(x) ((ll)(x).size())
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
template<class T = ll> using V = vector<T>;
template<class T = ll> using VV = V<V<T>>;
template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b; return true; }return false; }
template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b; return true; }return false; }
inline void Yes(bool b = true) {cout << (b ? "Yes" : "No") << '\n';}
inline void YES(bool b = true) {cout << (b ? "YES" : "NO") << '\n';}
inline void err(bool b = true) {if(b) {cout << -1 << '\n'; exit(0);}}
template<class T> inline void fin(bool b = true, T e = 0) {if(b) {cout << e << '\n'; exit(0);}}
template<class T> T Roundup_div(T x, T y) {return (x+(y-1))/y;}
const ll INF = 1e18;
template <typename T>

T pow(T a, long long n, int mod, T e = 1) {
    T ret = e;
    while (n) {
        if (n & 1) ret *= a;
        a *= a;
        ret %= mod;
        a %= mod;
        n >>= 1;
    }
    return ret % mod;
}

int solve(int n){
    if(n == 0) return 0;
    return solve(n % popcnt(n)) + 1;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll n; string s; cin >> n >> s;
    ll cnt = 0;
    REP(i, n) if(s[i] == '1') cnt++;
    ll psum = 0, msum = 0;
    REP(i, n) if(s[i] == '1'){
        psum += pow(2, n - i -1, cnt + 1);
        psum %= (cnt + 1);
    }
    if(cnt - 1) REP(i, n) if(s[i] == '1'){
        msum += pow(2, n - i - 1, cnt - 1);
        msum %= (cnt - 1);
    }
    REP(i, n){
        if(s[i] == '0'){
            ll x = psum + pow(2, n - i - 1, cnt + 1);
            x %= (cnt + 1);
            cout << solve(x) + 1 << endl;
        }
        else{
            if(!(cnt - 1)){
                cout << 0 << endl;
                continue;
            }
            ll y = msum - pow(2, n - i - 1, cnt - 1);
            y %= (cnt - 1);
            if(y < 0) y += (cnt - 1);
            cout << solve(y) + 1 << endl;
        }
    }
}