Equal Binary Subsequences
Everool has a binary string $$$s$$$ of length $$$2n$$$. Note that a binary string is a string consisting of only characters $$$0$$$ and $$$1$$$. He wants to partition $$$s$$$ into two disjoint equal subsequences. He needs your help to do it.You are allowed to do the following operation exactly once.   You can choose any subsequence (possibly empty) of $$$s$$$ and rotate it right by one position. In other words, you can select a sequence of indices $$$b_1, b_2, \ldots, b_m$$$, where $$$1 \le b_1 < b_2 < \ldots < b_m \le 2n$$$. After that you simultaneously set $$$$$$s_{b_1} := s_{b_m},$$$$$$ $$$$$$s_{b_2} := s_{b_1},$$$$$$ $$$$$$\ldots,$$$$$$ $$$$$$s_{b_m} := s_{b_{m-1}}.$$$$$$Can you partition $$$s$$$ into two disjoint equal subsequences after performing the allowed operation exactly once?A partition of $$$s$$$ into two disjoint equal subsequences $$$s^p$$$ and $$$s^q$$$ is two increasing arrays of indices $$$p_1, p_2, \ldots, p_n$$$ and $$$q_1, q_2, \ldots, q_n$$$, such that each integer from $$$1$$$ to $$$2n$$$ is encountered in either $$$p$$$ or $$$q$$$ exactly once, $$$s^p = s_{p_1} s_{p_2} \ldots s_{p_n}$$$, $$$s^q = s_{q_1} s_{q_2} \ldots s_{q_n}$$$, and $$$s^p = s^q$$$.If it is not possible to partition after performing any kind of operation, report $$$-1$$$. If it is possible to do the operation and partition $$$s$$$ into two disjoint subsequences $$$s^p$$$ and $$$s^q$$$, such that $$$s^p = s^q$$$, print elements of $$$b$$$ and indices of $$$s^p$$$, i. e. the values $$$p_1, p_2, \ldots, p_n$$$.
In the first test case, $$$b$$$ is empty. So string $$$s$$$ is not changed. Now $$$s^p = s_1 s_2 = \mathtt{10}$$$, and $$$s^q = s_3s_4 = \mathtt{10}$$$.In the second test case, $$$b=[3,5]$$$. Initially $$$s_3=\mathtt{0}$$$, and $$$s_5=\mathtt{1}$$$. On performing the operation, we simultaneously set $$$s_3=\mathtt{1}$$$, and $$$s_5=\mathtt{0}$$$.So $$$s$$$ is updated to 101000 on performing the operation.Now if we take characters at indices $$$[1,2,5]$$$ in $$$s^p$$$, we get $$$s_1=\mathtt{100}$$$. Also characters at indices $$$[3,4,6]$$$ are in $$$s^q$$$. Thus $$$s^q=100$$$. We are done as $$$s^p=s^q$$$.In fourth test case, it can be proved that it is not possible to partition the string after performing any operation.
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). Description of the test cases follows.The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$), where $$$2n$$$ is the length of the binary string.The second line of each test case contains the binary string $$$s$$$ of length $$$2n$$$.It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
For each test case, follow the following output format.If there is no solution, print $$$-1$$$. Otherwise,  In the first line, print an integer $$$m$$$ ($$$0 \leq m \leq 2n$$$), followed by $$$m$$$ distinct indices $$$b_1$$$, $$$b_2$$$, ..., $$$b_m$$$(in increasing order).  In the second line, print $$$n$$$ distinct indices $$$p_1$$$, $$$p_2$$$, ..., $$$p_n$$$ (in increasing order). If there are multiple solutions, print any.
import sys
R = sys.stdin.readline
for _ in range(int(R())):
    (n, s) = (int(R()), R().rstrip())
    ones = s.count('1')
    if ones & 1:
        print(-1)
        continue
    (temp, rot) = (0, [])
    for i in range(0, 2 * n, 2):
        if not s[i] == s[i + 1]:
            if int(s[i]) == temp:
                rot += (i + 1,)
            else:
                rot += (i + 2,)
            temp = 1 - temp
    print(len(rot), *rot)
    print(*range(1, 2 * n, 2))
n = int(input())
for _ in range(n):
    m = int(input())
    n = [int(x) for x in list(input())]
    if sum(n) % 2 == 1:
        print(-1)
    else:
        b = []
        ans = []
        have = 0
        for i in range(m):
            if n[2 * i] != n[2 * i + 1]:
                if n[2 * i] != have:
                    b.append(2 * i + 1)
                else:
                    b.append(2 * i + 2)
                have = (have + 1) % 2
            ans.append(2 * i + 1)
        print(len(b))
        b = ' '.join([str(x) for x in b])
        print(b)
        ans = ' '.join([str(x) for x in ans])
        print(ans)
for t in range(int(input())):
    n = int(input())
    a_str = input()
    a = [int(i) for i in a_str]
    d = []
    if sum(a) % 2 == 1:
        print(-1)
        continue
    for i in range(n):
        if a[2 * i] != a[2 * i + 1]:
            if len(d) % 2 == a[2 * i]:
                d.append(2 * i)
            else:
                d.append(2 * i + 1)
    print(len(d), end=' ')
    for i in range(len(d)):
        print(d[i] + 1, end=' ')
    print()
    for i in range(n):
        print(2 * i + 1, end=' ')
    print()