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: Wrong Answer)

fun main() = Thread(null, ::aising2020d, "(^o^)", 128 * 1024 * 1024).apply { start() }.join()

@OptIn(ExperimentalStdlibApi::class)
fun aising2020d() {
    val n = readLine()!!.toInt()
    val x = readLine()!!

    var bitCount = 0
    for (c in x) if (c == '1') bitCount++

    val bitCountToModded = mutableMapOf<Int, Long>()

    for (bc in bitCount - 1..bitCount + 1) {
        var modByN = 0L
        for ((index, c) in x.reversed().withIndex()) {
            if (c == '1') {
                modByN += (2L).modPow(index.toLong(), bc.toLong())
                modByN %= bc
            }
        }
        bitCountToModded[bc] = modByN
    }

    val answer = StringBuilder()

    for (i in 0 until n) {
        if (x[i] == '0') bitCount++ else bitCount--

        val modPow = (2L).modPow(n - i - 1L, bitCount.toLong())
        var modded = bitCountToModded[bitCount]!! + if (x[i] == '0') modPow else -modPow
        modded += bitCount
        modded %= bitCount

        if (modded == 0L) {
            answer.appendln(1)
            if (x[i] == '0') bitCount-- else bitCount++
            continue
        }

        var currX = modded.toInt()
        var count = 1

        while (currX > 0) {
            currX %= Integer.bitCount(currX)
            count++
        }

        answer.appendln(count)
        if (x[i] == '0') bitCount-- else bitCount++
    }

    print(answer)
}

private fun Long.modPow(p: Long, m: Long): Long {
    var x = this % m
    var y = p
    var result = 1L
    while (y > 0) {
        if (y % 2 == 1L) result = (result * x) % m
        y = y shr 1
        x = (x * x) % m
    }
    return result
}