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 PIGRECO 3.141592653589793

#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
  enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return rge<c> {i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
	~debug() { cerr << endl; }
	eni( != ) cerr << boolalpha << i; ris;
}
eni( == ) ris << range(begin(i), end(i));
}
sim, class b dor(pair < b, c > d) {
	ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
	*this << "[";
	for (auto it = d.b; it != d.e; ++it)
		*this << ", " + 2 * (it == d.b) << *it;
	ris << "]";
}
};
#define imie(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "
///////////////////////////////////////////////////////////////////////////
//////////////////// DO NOT TOUCH BEFORE THIS LINE ////////////////////////
///////////////////////////////////////////////////////////////////////////


#define int long long

int power(int x, unsigned int y, int mod) // binary exponentiation with modulo
{
	int res = 1;

	x = x % mod;


	if (x == 0)
		return 0;

	while (y > 0)
	{

		if (y & 1)
			res = (res * x) % mod;


		y = y >> 1;
		x = (x * x) % mod;
	}
	return res;
}



signed main() {

	ios::sync_with_stdio(false);

	cin.tie(0);


	int n;
	cin >> n;

	string s;
	cin >> s;

	int cnt = 0;

	for (int i = n - 1; i >= 0; i--) {
		if (s[i] == '1')
			cnt++;

	}



	long long val_piu = 0;
	long long val_meno = 0;


	reverse(s.begin(), s.end());

	for (int i = 0; i < n; i++) {
		if (s[i] == '1') {

			val_piu = val_piu + power(2, i, cnt + 1);


			val_meno = val_meno + power(2, i, cnt - 1);




		}
	}


	long long temp;



	for (int i = n - 1; i >= 0; i--) {
		int ans = 1;

		if (s[i] == '1') {

			temp = (val_meno - (power(2, i, cnt - 1))) % (cnt - 1);



		}

		else {

			temp = (val_piu + (power(2, i, cnt + 1))) % (cnt + 1);


		}


		while (temp) {
			temp = temp % (__builtin_popcountll(temp));
			ans++;

		}

		cout << ans << endl;


	}














}