This is the hard version of the problem. The difference between the versions is the constraint on $n$ and the required number of operations. You can make hacks only if all versions of the problem are solved.

There are two binary strings $a$ and $b$ of length $n$ (a binary string is a string consisting of symbols $0$ and $1$). In an operation, you select a prefix of $a$, and simultaneously invert the bits in the prefix ($0$ changes to $1$ and $1$ changes to $0$) and reverse the order of the bits in the prefix.

For example, if $a=001011$ and you select the prefix of length $3$, it becomes $011011$. Then if you select the entire string, it becomes $001001$.

Your task is to transform the string $a$ into $b$ in at most $2n$ operations. It can be proved that it is always possible.


-----Input-----

The first line contains a single integer $t$ ($1\le t\le 1000$)  — the number of test cases. Next $3t$ lines contain descriptions of test cases.

The first line of each test case contains a single integer $n$ ($1\le n\le 10^5$)  — the length of the binary strings.

The next two lines contain two binary strings $a$ and $b$ of length $n$.

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$.


-----Output-----

For each test case, output an integer $k$ ($0\le k\le 2n$), followed by $k$ integers $p_1,\ldots,p_k$ ($1\le p_i\le n$). Here $k$ is the number of operations you use and $p_i$ is the length of the prefix you flip in the $i$-th operation.


-----Example-----
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1

Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1



-----Note-----

In the first test case, we have $01\to 11\to 00\to 10$.

In the second test case, we have $01011\to 00101\to 11101\to 01000\to 10100\to 00100\to 11100$.

In the third test case, the strings are already the same. Another solution is to flip the prefix of length $2$, which will leave $a$ unchanged.

Difficulty: competition

Starter code:

Solutions:
import sys
input = sys.stdin.readline
from collections import deque

t=int(input())
for tests in range(t):
    n=int(input())

    a=input().strip()
    b=input().strip()


    Q=deque(a)

    L=[]
    while Q:
        L.append(Q.popleft())

        if Q:
            L.append(Q.pop())

    ANS=[]
    for i in range(n):
        if i%2==0:
            if L[i]==b[-1-i]:
                ANS.append(1)
        else:
            if L[i]!=b[-1-i]:
                ANS.append(1)

        ANS.append(n-i)

    print(len(ANS),*ANS)
            

    


for _ in range(int(input())):
    n = int(input())
    a = input()
    b = input()
    ans = []
    sign = -1
    for ansi in range(n, 0, -1):
        if (a[(n + ansi * sign) // 2] == b[ansi - 1]) ^ (sign > 0):
            ans.append(1)
        ans.append(ansi)
        sign = -sign
    print(len(ans), *ans)