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;
//Optimisations
#pragma GCC target ("avx2")
#pragma GCC optimization ("unroll-loops")
#pragma GCC optimize("O2")
//shortcuts for functions
#define pb push_back
#define mp make_pair
#define ff first 
#define ss second
#define all(v) v.begin(),v.end()
#define prec(n) fixed<<setprecision(n)
#define n_l '\n'
// make it python 
#define gcd __gcd
#define append push_back
#define str to_string
// utility functions shortcuts
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define sswap(a,b) {a=a^b;b=a^b;a=a^b;}
#define swap(a,b) {auto temp=a; a=b; b=temp;}
#define init(dp) memset(dp,-1,sizeof(dp));
#define set0(dp) memset(dp,0,sizeof(dp));
#define bits(x) __builtin_popcount(x)
#define SORT(v) sort(all(v))
#define endl "\n"
#define forr(i,n) for(ll i=0;i<n;i++)
// declaration shortcuts
typedef long long int ll;
#define int ll
// Constants
constexpr int dx[] = {-1, 0, 1, 0, 1, 1, -1, -1};
constexpr int dy[] = {0, -1, 0, 1, 1, -1, 1, -1};
constexpr ll INF = 1999999999999999997; 
constexpr int inf= INT_MAX;
constexpr int MAXSIZE = int(1e6)+5;
constexpr auto PI  = 3.14159265358979323846L;
constexpr auto oo  = numeric_limits<int>::max() / 2 - 2;
constexpr auto eps = 1e-6;
constexpr auto mod = 1000000007;
constexpr auto MOD = 1000000007;
constexpr auto MOD9 = 1000000009;
constexpr auto maxn = 100006;
//void IOfile(){
//freopen(file_name, reade_mode, stdin);
//freopen(file_name, write_mode, stdout);
//}
void fastio(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}

ll mod_pow(ll a,ll b,ll p){ //(a^b)%p
	ll res=1;
	a%=p;

	while(b>0){
		if(b&1==1) res=((res%p)*(a%p))%p;
		a=(a%p*a%p)%p;
		b>>=1;
	}
	return res;
}

//mod_inv=mod_pow(denominator,MOD-2,MOD);

int32_t main(){
	fastio();
	int n;
	cin >> n;
	int cnt = 0;
	int val = 0;
	string s;
	cin >> s;
	for(int i = 0; i < n; i++){
		if(s[i] == '1') cnt++;
		
	}
	if(cnt == 0){
		for(int i = 1; i <= n; i++){
			cout << 1 << "\n";
		}
		return 0;
	}
	if(cnt == 1){
		for(int i = 0 ; i < n; i++){
			if(s[i] == '1'){
				cout << 0 <<"\n";
			}else{
				if(s[n - 1] == '1')
				cout << 2 <<"\n";
				else cout << 1 <<"\n"; 
			}
		}
	}else{
		int x = cnt + 1, y = cnt - 1 ;
		int l = 0, r = 0;
		for(int i = 0 ; i < n; i++){
			if(s[i] == '1'){
				l += mod_pow(2, n - 1 - i, x);
				r += mod_pow(2, n - i - 1 , y);
			}
		}
		for(int i = 0; i < n; i++){
			if(s[i] == '1'){
				int to_minus = mod_pow(2,(n - i - 1) , y);
				int k = (r - to_minus + y + y) % y;
				int ans = 1;
				while(k > 0 ){
					k %= __builtin_popcount(k);
					ans++;
				}
				cout << ans <<"\n";
			}else{
				int to_add = mod_pow(2, n - i - 1 , x);
				int k = (l + to_add) % x;
				int ans = 1;
				while(k > 0 ){
					k %= __builtin_popcount(k);
					ans++;
				}
				cout << ans <<"\n";
			}
		}
	}
}