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.math.BigDecimal
import java.math.BigInteger

fun main() {
    val N = readLine()!!.toInt()
    val x = readLine()!!

    for (i in 0 until N) {
        println(calc(x, i, N))
    }
}

fun calc(x: String, i: Int, N: Int): BigInteger {
    var replaced = x.toCharArray()
    replaced[i] = if (replaced[i] == '0') '1' else '0'

    replaced = replaced.reversedArray()
    var target = BigInteger.ZERO
    for (i in 0 until N) {
        val tmp = calcPow(i.toBigInteger())
        if (replaced[i] == '1') {
            target += tmp
        }
    }
    return calcF(target)
}

fun calcF(i: BigInteger): BigInteger {
    if (i == BigInteger.ZERO) {
        return BigInteger.ZERO
    }
    val popCount = i.bitCount()
    val divided = i.mod(popCount.toBigInteger())
    return BigInteger.ONE + calcF(divided)
}

fun calcPow(i: BigInteger): BigInteger {
    if (i == BigInteger.ZERO) {
        return BigInteger.ONE
    }
    return BigInteger("2") * calcPow(i.minus(BigInteger.ONE))
}