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;

const int N = 200005;

int dp[N];
int n, pw[2][N], cnt[2], r[2];
char s[N];

int calc(int x) {
	return dp[x];
}

int main() {
	dp[0] = 0;
	for (int i = 1; i < N; i ++) dp[i] = dp[i%(__builtin_popcount(i))] + 1;
	cin >> n >> s;
	int c = 0;
	for (int i = 0; i < n; i ++) c += (s[i] == '1');
	cnt[0] = c + 1, cnt[1] = c - 1;
	for (int i = 0; i < 2; i ++) {
		pw[i][0] = 1;
		for (int j = 1; j < n; j ++) {
			pw[i][j] = pw[i][j-1] << 1;
			if (pw[i][j] >= cnt[i]) pw[i][j] -= cnt[i];
		}
		r[i] = 0;
		for (int j = 0; j < n; j ++) {
			r[i] = r[i] * 2 + (s[j] - '0');
			if (r[i] >= cnt[i]) r[i] -= cnt[i];
		}
	}
	for (int i = 0; i < n; i ++) {
		if (s[i] == '1') {
			int c = cnt[1];
			if (!c) cout << 0 << endl;
			else {
				int R = r[1] - pw[1][n-1-i];
				if (R >= cnt[1]) R -= cnt[1];
				cout << calc(R) + 1 << endl;
			}
		}
		else {
			int R = r[0] + pw[0][n-1-i];
			if (R >= cnt[0]) R -= cnt[0];
			cout << calc(R) + 1 << endl;
		}
	}
	return 0;
}