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>
using namespace std;

long long M;

long long BigMod(long long b, long long m) {
	if(m == 0) return 0;
	long long a = 2;
    a %= m;
    long long res = 1;
    while (b > 0) {
        if (b & 1)
            res = (res * a) % m;
        a = (a * a) % m;
        b >>= 1;
    }
    return res;
}

long long f(long long n)
{
	long long step = 0;
	while(__builtin_popcount(n)) {
		step++;
		int x = __builtin_popcount(n);
		n %= x;
	}
	return step;
}

int main()
{
	int n;
	scanf("%d", &n);
	string x;
	cin >> x;
	long long num_of_one = 0;
	for(int i = 0; i < x.size(); i++) {
		num_of_one += x[i] == '1';
	}
	reverse(x.begin(), x.end());

	long long mod_zero[300000];
	long long mod_one[300000];
	memset(mod_zero, 0, sizeof(mod_zero));
	memset(mod_one, 0, sizeof(mod_one));
	for(int i = 0; i < n; i++) {
		if(x[i] == '1') {
			mod_zero[i] = BigMod(i, num_of_one+1);
			mod_one[i] = BigMod(i, num_of_one-1);
		}
		if(i > 0) {
			mod_zero[i] += mod_zero[i-1];
			mod_zero[i] %= (num_of_one+1);
			mod_one[i] += mod_one[i-1];
			if(num_of_one > 1) {
				mod_one[i] %= (num_of_one-1);
			}
		}
	}
	long long m_1 = mod_zero[n-1];
	long long m_2 = mod_one[n-1];

	long long les = num_of_one-1;
	long long gre = num_of_one+1;

	long long ans[300000];
	for(int i = 0; i < n; i++) {
		if(x[i] == '1') {
			if(les > 0) {
				long long k = BigMod(i, les);
				long long z = (m_2-k)%les;
				if(z < 0) z += les;
				ans[i] = 1 + f(z);
			} else {
				ans[i] = 0;
			}

		} else {
			long long k = BigMod(i, gre);
			long long z = (m_1+k) % gre;
			ans[i] = 1 + f(z);
		}
	}
	for(int i = n-1; i>=0; i--) {
		printf("%lld\n", ans[i]);
	}
}