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 <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <fstream>
#include <cassert>
#include <cstring>
#include <unordered_set>
#include <unordered_map>
#include <numeric>
#include <ctime>
#include <bitset>
#include <complex>
#include <chrono>
#include <random>
#include <functional>

using namespace std;

#define int long long

int f(int x) {
	int ans = 0;
	while (x != 0) {
		ans++;
		x %= (__builtin_popcount(x));
	}
	return ans;
}


int bp(int a, int b, int m) {
	int res = 1;
	while (b) {
		if (b & 1) {
			res *= a;
			res %= m;
		}
		a *= a;
		a %= m;
		b >>= 1;
	}
	return res;
}

signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	int n;
	cin >> n;
	string s;
	cin >> s;
	int cnt = 0;
	for (int i = 0; i < n; i++) {
		cnt += (s[i] == '1');
	}
	int modplus = 0;
	int modmius = 0;
	for (int i = 0; i < n; i++) {
		modplus = (2 * modplus + (s[i] - '0')) % (cnt + 1);
		if (cnt > 1) modmius = (2 * modmius + (s[i] - '0')) % (cnt - 1);
	}
	for (int i = 0; i < n; i++) {
		if (s[i] == '0') {
			cout << 1 + f((modplus + bp(2, n - 1 - i, cnt + 1)) % (cnt + 1)) << '\n';
		} else {
			if (cnt == 1) {
				cout << 0 << '\n';
			} else {
				int v = modmius - bp(2, n - 1 - i, cnt - 1);
				v %= (cnt - 1);
				v += (cnt - 1);
				v %= (cnt - 1);
				cout << 1 + f(v) << '\n';
			}
		}
	}
}