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++ (Clang 10.0.0) (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<set>
#include<string>
typedef long long ll;
using namespace std;
#define INF (1 << 30)
#define INFLL (1LL << 60)
#define rep(i, n) for(int i=0; i<(n); ++i)


int popcount(string bin)
{
	int cnt = 0;
	rep(i, bin.size()) if (bin[i] == '1') ++cnt;
	return cnt;
}
int popcount(ll bits)
{
	bits = (bits & 0x55555555) + (bits >> 1 & 0x55555555);
	bits = (bits & 0x33333333) + (bits >> 2 & 0x33333333);
	bits = (bits & 0x0f0f0f0f) + (bits >> 4 & 0x0f0f0f0f);
	bits = (bits & 0x00ff00ff) + (bits >> 8 & 0x00ff00ff);
	return (bits & 0x0000ffff) + (bits >> 16 & 0x0000ffff);
}
int funcOnce(string bin, int div)
{
	int deci = 0;
	rep(i, bin.size()) deci = (deci * 2 + bin[i] - '0') % div;
	return deci;
}
int funcOnce(int deci, int div) { return deci % div; }

int func(string bin, int div)
{
	if (div == 0) return 0;
	int deci = funcOnce(bin, div);
	int bitcnt = popcount(deci);
	int cnt = 1;
	while (deci > 0) {
		deci = funcOnce(deci, bitcnt);
		bitcnt = popcount(deci);
		++cnt;
	}
	return cnt;
}

int main()
{
	int N; cin >> N;
	string X; cin >> X;
	int bitcnt = popcount(X);
	int xmodp1 = funcOnce(X, bitcnt + 1);
	int xmodm1;
	if (bitcnt > 1) {
		xmodm1 = funcOnce(X, bitcnt - 1);
	}

	vector<int> ans(N, 0);

	rep(i, N) {
		string ibin(N - i, '0');
		ibin[0] = '1';
		int mod, imod, xmod;
		if (X[i] == '1') {
			mod = bitcnt - 1;
			if (mod < 1) continue;
			imod = funcOnce(ibin, mod);
			xmod = (mod + xmodm1 - imod) % mod;
		}
		else {
			mod = bitcnt + 1;
			imod = funcOnce(ibin, mod);
			xmod = (xmodp1 + imod) % mod;
		}
		++ans[i];
		while (xmod > 0) {
			xmod = funcOnce(xmod, popcount(xmod));
			++ans[i];
		}
	}

	rep(i, N) cout << ans[i] << endl;
	return 0;
}