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 INF 1000000007
#define LINF (1LL << 62)
#define PI 3.14159265358979

typedef long long i64;
typedef pair<i64,i64> P;

inline i64 mod(i64 a, i64 m) { return (a % m + m) % m; }

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }



int n;
string s;
int dp[202020], dp2[202020];
void solve(){
	cin >> n >> s;
	
	int cnt = 0;
	for(int i = 0; i < n; i++){
		if(s[i] == '1') cnt++;
	}
	
	int v = 1, v2 = 1;
	for(int i = 0; i < n; i++){
		if(cnt > 1){
			v %= cnt-1;
			dp[n-i-1] = v;
			v *= 2;
		}
		v2 %= cnt+1;
		dp2[n-i-1] = v2;
		v2 *= 2;
	}
	
	i64 sum = 0, sum2 = 0;
	for(int i = 0; i < n; i++){
		if(s[i] == '1'){
			if(cnt > 1){
				sum += dp[i];
				sum %= cnt-1;
			}
			sum2 += dp2[i];
			sum2 %= cnt+1;
		}
	}
	
	
	for(int i = 0; i < n; i++){
		i64 a = 0;
		if(s[i] == '1'){
			if(cnt > 1) a = mod(sum-dp[i], cnt-1);
			else{
				cout << 0 << endl;
				continue;
			}
		}
		else{
			a = mod(sum+dp2[i], cnt+1);
		}
		int ans = 1;
		while(a != 0){
			ans++;
			int cc = 0;
			for(int j = 0; (1LL<<j) < a*2; j++){
				if(a>>j & 1) cc++;
			}
			a %= cc;
		}
		cout << ans << endl;
	}
	
}

int main(){
	std::cin.tie(0);
	std::ios::sync_with_stdio(false);
	int t = 1;
	//cin >> t;
	while(t--){
		solve();
	}
	
	return 0;
}