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, s, n) for (int i = (s); i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define debug(x) cout<<#x<<": "<<x<<endl
using namespace std;
using ll = long long;
const ll MOD = 1e9+7;
const ll LINF = LLONG_MAX;
const int INF = INT_MAX;
const double PI=acos(-1);
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 main(){
	ll n; cin>>n;
	string x; cin>>x;
	ll pc_0 = 0;
	rep(i,0,x.size()){
		if(x[i]=='1') pc_0++;
	}
	ll m = pc_0 - 1;
	ll p = pc_0 + 1;
	vector<ll> mod_m(200010, 0);
	vector<ll> mod_p(200010, 0);
	ll mm = 1;
	// mod_m[i] に2^iをわったあまりがはいっている
	rep(i,0,200010){
		ll temp = mm % m;
		mod_m[i] = temp;
		mm = temp * 2;
	}
	ll pp = 1;
	rep(i,0,200010){
		ll temp = pp % p;
		mod_p[i] = temp;
		pp = temp * 2;
	}

	vector<vector<ll>> mod_main(25, vector<ll>(25, 0));
	rep(mod,1,25){
		ll now = 1;
		rep(i,0,25){
			ll temp = now % mod;
			mod_main[mod][i] = temp;
			now = temp * 2;
		}
	}

	// 前処理終わり

	for(int rev = 0; rev < x.size(); rev++){
		ll res = 0;
		// 反転
		string now = x;
		ll popcnt = pc_0;
		char flag;
		if(x[rev] == '0'){
			now[rev] = '1';
			popcnt++;
			flag = 'p';
		}else{
			now[rev] = '0';
			popcnt--;
			flag = 'm';
		}

		// 1回目
		ll amari_1 = 0;
		for(int i = x.size() - 1; i >= 0; i--){
			if(now[i] == '1'){
				if(flag = 'p'){
					amari_1 += mod_p[i];
				}else{
					amari_1 += mod_m[i];
				}
			}
		}
		amari_1 %= popcnt;
		res++;
		if(amari_1 == 0){
			cout<<res<<endl;
			continue;
		}

		// 2回目以降
		ll y = amari_1;

		while(y > 0){
			ll amari = 0;
			ll pcnt = __builtin_popcount(y);
			for(int bit = 0; bit <= 24; bit++){
				if((1 << bit) & y) amari += mod_main[pcnt][bit];
			}
			amari %= pcnt;
			y = amari;
			res++;
		}
		cout << res << endl;
	}

}