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 <iostream>
#include <algorithm>
#include <string>
#include <queue>
#include <cmath>
#include <map>
#include <iomanip>
#include <unordered_map>
#include <stack>
#include <cstring>
#include <cctype>
#include <random>

using namespace std;
typedef long long  ll;
const ll INF = 1000000007;
const ll mod = 998244353;

vector<ll> dp(2 * 100000 + 1, -1);
vector<ll> popcount(2 * 100000 + 1, 0);

int pcnt(int x) {
	return __builtin_popcount(x);
}

int f(int x) {
	if (x == 0) return 0;
	return f(x % pcnt(x)) + 1;
}

int main()
{
	ll n;
	cin >> n;
	string s;
	cin >> s;
	vector<ll> x(n);
	// 0,1変換
	for (ll i = 0; i < n; i++)
	{
		x[i] = s[i] - '0';
	}
	// xのpopcount
	ll pc = 0;
	for (ll i = 0; i < n; i++)
	{
		pc += x[i];
	}
	vector<ll> ans(n);
	for (ll b = 0; b < 2; b++)
	{
		ll npc = pc;
		if (b == 0) npc++; else npc--;
		if (pcnt) continue;
		// xのnpc
		ll r0 = 0;
		for (ll i = 0; i < n; i++)
		{
			r0 = (r0 * 2) % npc;
			r0 += x[i];
		}
		ll k = 1;
		for (ll i = n - 1; i >= 0; --i) {
			if (x[i] == b) {
				int r = r0;
				if (b == 0) r = (r + k) % npc;
				else r = (r - k + npc) % npc;
				ans[i] = f(r) + 1;
			}
			k = (k * 2) % npc;
		}
	}

	for (ll i = 0; i < n; i++)
	{
		cout << ans[i] << endl;
	}
	return 0;
}