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 fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long ll;
typedef pair<ll,ll> P;
const ll mod=1000000007;
const ll LINF=1LL<<60;
const int INF=1<<30;
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; }

//繰り返し二乗法
ll mod_pow(ll x,ll n,ll p){
	ll res = 1;
	while(n > 0){
		if(n & 1){
			res = res * x % p;
		}
		x = x * x % p;
		n >>= 1;
	}
	return res;
}

int main(){
	ll n;cin>>n;
	string x;cin>>x;
	ll bit = 0;
	for(ll i = 0; i < n; i++) {
		if (x[i] == '1'){
			bit++;
		}
	}
	vector<ll> pow1(n + 1,1);
	vector<ll> pow2(n + 1,1);
	if(x[n - 1] == '0'){
		pow1[0] = 0;
		pow2[0] = 0;
	}
	for(ll i = 1; i < n; i++) {
		if (x[n - 1- i] == '1'){
			pow1[i] = (mod_pow(2,i,bit + 1) + pow1[i - 1]) % (bit + 1);
			if(bit - 1 != 0){
				pow2[i] = (mod_pow(2,i,bit - 1) + pow2[i - 1]) % (bit - 1);
			}
			else{
				pow2[i] = 0;
			}
		}
		else{
			pow1[i] = pow1[i - 1];
			pow2[i] = pow2[i - 1];
		}
		
	}
	// for(int i = 0; i < n; i++) {
	// 	cout << pow1[i] << endl;
	// }
	// cout << "bit" << bit << endl;
	for(ll i = 0; i < n; i++) {
		if (x[i] == '1'){
			if(bit - 1 == 0){
				cout << 1 << endl;
				return 0;
			}
			ll tmp = (pow2[n - 1] - (mod_pow(2,n - 1 - i,bit - 1))) % (bit - 1);
			ll ans = 1;
			if(tmp < 0){
				tmp +=(bit - 1);
			}
			// cout << "tmp" << tmp << endl;
			while(tmp != 0){
				ans++;
				ll div = __builtin_popcount(tmp);
				// cout << "div" << div << endl;
				// cout << "tmp" << tmp << endl;
				tmp %= div;
			}
			cout << ans << endl;
		}
		else{
			ll tmp = (pow1[n - 1] + (mod_pow(2,n - 1 - i,bit + 1))) % (bit + 1);
			ll ans = 1;
			// cout << "tmp" << tmp << endl;
			while(tmp != 0){
				ans++;
				ll div = __builtin_popcount(tmp);
				// cout << "div" << div << endl;
				// cout << "tmp" << tmp << endl;
				tmp %= div;
			}
			cout << ans << endl;

		}
		
	}
	return 0;
}