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: Runtime Error)

import java.util.*

fun main(args: Array<String>) {
    val pw = java.io.PrintWriter(System.out)
    Problem.solve(Scanner(System.`in`), pw)
    pw.flush()
}

object Problem {
    fun next(X: IntArray, div: Int): Int {
        var ans = 0
        var d = 1
        for (i in X.count() - 1 downTo 0) {
            if (X[i] == 1) {
                ans += d % div
                ans %= div
            }
            d *= 2
            d %= div
        }
        return ans
    }

    fun bits(n: Int): Int {
        var i = n
        var b = 0
        do {
            b += i and 1
            i /= 2
        } while (i > 0)
        return b
    }

    fun next(X: Int): Int {
        var ans = 0
        var d = 1
        val div = bits(X)
        var x = X
        while (x > 0) {
            if (x and 1 == 1) {
                ans += d % div
                ans %= div
            }
            d *= 2
            d %= div
            x /= 2
        }
        return ans
    }

    fun solve(sc: Scanner, pw: java.io.PrintWriter) {
        val N = sc.next().toInt()
        val X = sc.next().map { it.toInt() - 48 }.toIntArray()

        val n = IntArray(100001)
        for (i in 1..100000) {
            n[i] = next(i)
        }

        val div = X.count { it == 1 }
        val x1 = next(X, div - 1)
        val x2 = next(X, div + 1)
        val second = IntArray(N)
        var d = 1
        for (i in N - 1 downTo 0) {
            if (X[i] == 1) {
                // 1 -> 0
                second[i] = x1 - d % (div - 1)
                second[i] += div - 1
                second[i] %= div - 1
            } else {
                // 0 -> 1
                second[i] = x2 + d % (div + 1)
                second[i] %= div + 1
            }
            d *= 2
        }

        for (i in 0 until N) {
            var ans = 1
            var x = second[i]
            while (x > 0) {
                x = n[x]
                ans++
            }
            pw.println(ans)
        }
    }
}