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: Accepted)

#include <bits/stdc++.h>
#define PREP(i, s, x) for(ll i = (s); i < (x); i++)
#define MREP(i, s, x) for(ll i = (s); i >= (x); i--)
#define REP(i, x) PREP(i, 0, x)
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template<typename T> inline bool chmax(T &a, T b) { return ((a < b) ? (a = b, true) : (false)); }
template<typename T> inline bool chmin(T &a, T b) { return ((a > b) ? (a = b, true) : (false)); }

// variable -------------------------------------------

constexpr ll INF = (1ll << 59);
constexpr ll MOD = 1000000007ll;

// structure ------------------------------------------

// function -------------------------------------------

ll amari(string x, ll m){
	ll res = 0;
	ll n = x.length();
	ll prev = 0;
	REP(i, n){
		ll p = x[i] - '0';
		prev = (p + prev * 2) % m;
	}
	res = prev;
	return res;
}

ll popo(string x){
	ll res = 0;
	ll n = x.length();
	REP(i, n){
		if(x[i] == '1'){
			res ++;
		}
	}
	return res;
}

ll popo(ll x){
	// 10->2
	string s = "";
	while(x != 0){
		s += char((x % 2) + '0');
		x /= 2;
	}
	reverse(s.begin(), s.end());
	return popo(s);
}

// main -----------------------------------------------

int main() {
	ll n;
	string x;
	cin >> n >> x;
	ll memo1, memo2 = 0;
	ll pp = popo(x);
	ll yabaiwayo = -1;
	if(pp == 1){
		REP(i, n){
			if(x[i] == '1'){
				yabaiwayo = i;
				break;
			}
		}
	}else{
		memo2 = amari(x, pp - 1);
	}
	memo1 = amari(x, pp + 1);
	// cout << memo1 << memo2 << pp <<  endl;

	vector<ll> ans(n);
	ll nibeki1 = 1, nibeki2 = 1;
	// 下の桁から見ている
	REP(i, n){
		// x, ans専用
		ll index = n - i - 1;
		ll mini_ans, cnt = 0;
		if(yabaiwayo != -1 && yabaiwayo == index){
			ans[index] = 0;
			nibeki1 *= 2;
			nibeki1 %= (pp + 1);
			// cout << "y";
			continue;
		}
		// cout << x[index];
		if(x[index] == '1'){
			mini_ans = (((memo2 - nibeki2) % (pp - 1)) + (pp - 1)) % (pp - 1);
			// cout << "!" << endl;
		}else{
			mini_ans = (memo1 + nibeki1) % (pp + 1);
			// cout << "?" << endl;
		}
		nibeki1 *= 2;
		nibeki1 %= (pp + 1);
		if(pp - 1 != 0){
			nibeki2 *= 2;
			nibeki2 %= (pp - 1);
		}
		cnt ++;
		while(mini_ans != 0){
			mini_ans = mini_ans % popo(mini_ans);
			cnt ++;
		}
		ans[index] = cnt;
	}

	// cout << endl;

	REP(i, n){
		cout << ans[i] << endl;
	}
	return 0;
}