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;

#define ll long long int
#define f(i,x,n)  for(int i=x;i<n;i++)
#define FASTIO cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(false);
#define sz(a) int((a).size())
#define mod 1000000007
#define inf LLONG_MAX
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
template <typename T>
inline T gcd(T a, T b) { while (b != 0) swap(b, a %= b); return a; }

template <typename T>
inline void seethis(vector<T>vect)
{
    for(T x:vect)
        cout<<x<<" ";
    cout<<"\n";
}

void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
	cerr << *it << " = " << a << endl;
	err(++it, args...);
}
//LCM
int lcm(int a, int b) {  return a * (b / gcd(a, b)); }

//For sort by second
bool cmp(const pair<int,int> &a)
{ return a.first<a.second; }

//Tree
/* const int MAXN = 1e5 + 5;
vector<ll> adj[MAXN];*/

//ModPower
ll modpower(ll a,ll b,ll c)
{
	ll res=1;
	while(b)
	{
		if(b&1LL)
			res=(res*a)%c;
		a=(a*a)%c;
		b>>=1;
	}
	return res;
}

ll popcount(ll n)
{
	ll ans=0;
	ll bits=__builtin_popcount(n);
	while(n)
	{
		n=n%bits;
		ans++;
		bits=__builtin_popcount(n);
	}
	return ans;
}
ll getMod(string str, ll n, ll k)  
{  
  
    // pwrTwo[i] will store ((2^i) % k)  
    ll pwrTwo[n];  
    pwrTwo[0] = 1 % k;  
    for (ll i = 1; i < n; i++)  
    {  
        pwrTwo[i] = pwrTwo[i - 1] * (2 % k);  
        pwrTwo[i] %= k;  
    }  
  
    // To store the result  
    ll res = 0;  
    ll i = 0, j = n - 1;  
    while (i < n)  
    {  
  
        // If current bit is 1  
        if (str[j] == '1')  
        {  
  
            // Add the current power of 2  
            res += (pwrTwo[i]);  
            res %= k;  
        }  
        i++;  
        j--;  
    }  
    return res;  
}  

ll solve(string s,int n,int bits)
{
	ll temp= getMod(s,n,bits);
	ll ans=1;
	bits=__builtin_popcount(temp);
	while(temp)
	{
		temp=temp%bits;
		ans++;
		bits=__builtin_popcount(temp);
	}
	return ans;
}

int main()
{
	FASTIO;
	#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	#endif
	ll n;
	 cin>>n;

	 string s;
	  cin>>s;
	  
	 
	  ll bits=0;
	 for(int i=0;i<n;i++)
	 {
	 	if(s[i]=='1')
	 		bits++;
	 }
 	vector<ll>ans;


	 for(int i=0;i<n;i++)
	 {
	 	if(s[i]=='0')
	 	{
	 		s[i]='1';
	 		ll x= solve(s,n,bits+1);
	 		s[i]='0';
	 		ans.push_back(x);
	 	}
	 	else
	 	{
	 		s[i]='0';
	 		ll x= solve(s,n,bits-1);
	 		ans.push_back(x);
	 		s[i]='1';
	 	}
	 }


	 for(int i=0;i<n;i++)
	 {
	 	cout<<ans[i]<<"\n";
	 }

	
	return 0;
}