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>
using namespace std;

const int N = 2e5+7;
typedef long long LL;

int n,sum;
string s;

int quick(int a,int b,int mod)
{
	LL ans=1;
	while(b){
		if(b&1) ans = ans*a%mod;
		a = (LL)a*a%mod;
		b>>=1;
	}
	return ans;
}

int fun(int n){
	int ans=0;
	while(n){
		ans++;
		n  = n &(n-1);
	} return ans;
}

int main()
{
	cin>>n>>s;
	for(auto c :s) if(c=='1') sum++;
	int cnt1 = 0 ,cnt2 = 0, p1 = sum + 1 , p2 = sum - 1;
	for(int i=0;i<n;i++)
		if(s[i]=='1'){
			cnt1 = (cnt1 + quick(2,n-i-1,p1))%p1;
			cnt2 = (cnt2 + quick(2,n-i-1,p2))%p2;
		}
	int t,ans=0;
	for(int i=0;i<n;i++){
		ans=1;
		if(s[i]=='1') t = (cnt2-quick(2,n-i-1,p2)+p2)%p2;
		else  t = (cnt1+quick(2,n-i-1,p1)+p1)%p1;
		if(!t) cout<<ans<<"\n";
		else {
			while(t) ans++,t = t%fun(t);
			cout<<ans<<endl;
		}
	}
	return 0;
}