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 Java (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class Main {

    private static FastReader in = new FastReader(System.in);
    private static PrintWriter out = new PrintWriter(System.out);

    private static final long MOD = (long) 1e9 + 7;

    public static void main(String[] args) {
        solve();
        out.flush();
    }

    private static void solve() {
        int n = in.nextInt();
        boolean[] x = new boolean[n];
        char[] ch = in.next().toCharArray();
        for (int i = 0; i < ch.length; i++) {
            x[i] = ch[i] == '1';
        }

        int popCount = popCount(x);

        BigInteger bi = toBigInteger(x);
        int popCountMinus1 = Math.max(1, popCount - 1);
        int mod0 = bi.mod(BigInteger.valueOf(popCountMinus1)).intValue();
        int mod1 = bi.mod(BigInteger.valueOf(popCount + 1)).intValue();

        int[][] modulus = new int[2][n];
        modulus[0][n - 1] = 1 % (popCountMinus1);
        modulus[1][n - 1] = 1 % (popCount + 1);
        for (int i = n - 1; i >= 1; i--) {
            modulus[0][i - 1] = (modulus[0][i] * 2) % (popCountMinus1);
        }
        for (int i = n - 1; i >= 1; i--) {
            modulus[1][i - 1] = (modulus[1][i] * 2) % (popCount + 1);
        }

        for (int i = 0; i < n; i++) {
            if (x[i]) {
                if (popCount == 1) {
                    out.println(0);
                    continue;
                }
                out.println(f((mod0 - modulus[0][i] + popCountMinus1) % (popCountMinus1)) + 1);
            } else {
                out.println(f((mod1 + modulus[1][i]) % (popCount + 1)) + 1);
            }
        }
    }

    private static int f(int b) {
        int ret = 0;
        while (b != 0) {
            b = b % (Integer.bitCount(b));
            ret++;
        }
        return ret;
    }

    private static BigInteger toBigInteger(boolean[] x) {
        BigInteger bi = BigInteger.ZERO;
        BigInteger two = BigInteger.valueOf(2);
        for (int i = 0; i < x.length; i++) {
            if (!x[i]) continue;
            bi = bi.add(two.pow(x.length - i - 1));
        }
        return bi;
    }

    private static int popCount(boolean[] x) {
        int ret = 0;
        for (boolean ib : x) {
            if (ib) ret++;
        }
        return ret;
    }

    // https://www.geeksforgeeks.org/fast-io-in-java-in-competitive-programming/
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        FastReader(InputStream inputStream) {
            br = new BufferedReader(new
                    InputStreamReader(inputStream));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException  e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        // https://qiita.com/masakinpo/items/5f77f2a879578a85a0bb
        public int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }
    }
}