Problem:
Score : 400 points

Problem Statement

There are N empty boxes arranged in a row from left to right.
The integer i is written on the i-th box from the left (1 \leq i \leq N).

For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it.

We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied:

For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2.

Does there exist a good set of choices? If the answer is yes, find one good set of choices.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

a_i is 0 or 1.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

If a good set of choices does not exist, print -1.

If a good set of choices exists, print one such set of choices in the following format:

M
b_1 b_2 ... b_M

where M denotes the number of boxes that will contain a ball, and b_1,\ b_2,\ ...,\ b_M are the integers written on these boxes, in any order.

Sample Input 1

3
1 0 0

Sample Output 1

1
1

Consider putting a ball only in the box with 1 written on it.

There are three boxes with multiples of 1 written on them: the boxes with 1, 2, and 3. The total number of balls contained in these boxes is 1.

There is only one box with a multiple of 2 written on it: the box with 2. The total number of balls contained in these boxes is 0.

There is only one box with a multiple of 3 written on it: the box with 3. The total number of balls contained in these boxes is 0.

Thus, the condition is satisfied, so this set of choices is good.

Sample Input 2

5
0 0 0 0 0

Sample Output 2

0

Putting nothing in the boxes can be a good set of choices.

Solution in Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"fmt"
)

const MOD = 1000000007

func ModAdd(a, b int64) int64 {
	return (a + b) % MOD
}

func ModMul(a, b int64) int64 {
	return a * b % MOD
}

func ModSub(a, b int64) int64 {
	r := (a%MOD - b%MOD) % MOD
	if r < 0 {
		r += MOD
	}
	return r
}

func ModDiv(a, b int64) int64 {
	return a % MOD * ModInv(b) % MOD
}

func ModPow(a, n int64) int64 {
	var r int64 = 1
	for n > 0 {
		if n&1 == 1 {
			r = r * a % MOD
		}
		a = a * a % MOD
		n >>= 1
	}
	return r
}

func ModInv(a int64) int64 {
	var b, u, v int64 = MOD, 1, 0
	for b > 0 {
		t := a / b
		a -= t * b
		a, b = b, a
		u -= t * v
		u, v = v, u
	}
	u %= MOD
	if u < 0 {
		u += MOD
	}
	return u
}

type Q struct {
	idx, value int
	a, b       int
	index      int
}

type QS []*Q

func (p QS) Len() int           { return len(p) }
func (p QS) Less(i, j int) bool { return p[i].value > p[j].value }
func (pq QS) Swap(i, j int) {
	pq[i], pq[j] = pq[j], pq[i]
	pq[i].index = i
	pq[j].index = j
}

func (pq *QS) Push(x interface{}) {
	n := len(*pq)
	item := x.(*Q)
	item.index = n
	*pq = append(*pq, item)
}

func (pq *QS) Pop() interface{} {
	old := *pq
	n := len(old)
	item := old[n-1]
	old[n-1] = nil  // avoid memory leak
	item.index = -1 // for safety
	*pq = old[0 : n-1]
	return item
}

func FactoredPrimeNumber(num int) map[int]int {
	m := map[int]int{}
	return factor(m, num, 2)
}

func factor(result map[int]int, num, pnum int) map[int]int {
	if pnum*pnum > num {
		if num != 1 {
			result[num] += 1
		}
		return result
	}

	if num%pnum == 0 {
		num /= pnum
		result[pnum]++
	} else {
		pnum++
	}
	return factor(result, num, pnum)
}

func main() {
	var N int
	fmt.Scanf("%d\n", &N)
	A := make([]int, N)
	for i := 0; i < N; i++ {
		var a int
		fmt.Scan(&a)
		A[i] = a
	}
	B := make([]int, N)
	max := 0
	for i := N; i > 0; i-- {
		a := A[i-1]
		if a == 0 {
			continue
		}
		p := 2
		sum := 0
		for i*p <= N {
			a := A[i*p-1]
			if a != 0 {
				sum++
			}
			p++
		}
		if sum%2 == 0 {
			B[i-1] = 1
			if i > max {
				max = i
			}
		}
	}
	if max == 0 {
		fmt.Println(0)
	} else {
		fmt.Println(max)
		for i := 0; i < max; i++ {
			fmt.Print(B[i])
			fmt.Print(" ")
		}
		fmt.Println()
	}
}

func Reverse(runes []rune) []rune {
	for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
		runes[i], runes[j] = runes[j], runes[i]
	}
	return runes
}