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 Kotlin (1.3.71) (judge verdict: Time Limit Exceeded)

import java.io.PrintWriter

fun main() {
    val n = readLine()!!.toInt()
    val s = readLine()!!
    val x = s.toBigInteger(2)
    val bc = x.bitCount()
    val x1 = x % (bc + 1).toBigInteger()
    val x2 = if (bc - 1 > 0) x % (bc - 1).toBigInteger() else 0.toBigInteger()

    val pw = PrintWriter(System.out)

    for (i in (0 until n)) {
        val b = 2.toBigInteger().pow(n - i - 1)
        var cnt = 0
        var xi = if (s[i] == '1') {
            if (bc - 1 == 0) {
                0.toBigInteger()
            } else {
                cnt++
                val mod = (bc - 1).toBigInteger()
                (x2 + mod - (b % mod)) % mod
            }
        } else {
            cnt++
            val mod = (bc + 1).toBigInteger()
            (x1 + (b % mod)) % mod
        }.toInt()

        while (xi > 0) {
            cnt++
            xi %= Integer.bitCount(xi)
        }

        pw.println(cnt)
    }

    pw.flush()
}