You are given two binary strings $s$ and $t$, each of the same length $n$. You are allowed to perform the following operation:
- Pick indices
$l$,$r$($1 \le \color{red}{l \lt r} \le n$) such that substring$s_{l,r}$is a palindrome and flip all bits in substring$s_{l,r}$.
The goal is to finally make $s$ equal to $t$, performing any of the above operations at most $2n$ times (possibly none).
A substring $s_{l,r}$ of a string $s$ is the contiguous sequence of characters starting from index $l$ and ending at index $r$ (both inclusive), where $1 \leq l \lt r \leq |s|$. Here $|s|$ denotes the length of the string $s$.
A string is a palindrome if it reads the same forwards and backwards. For example, the strings 101 and 00 are palindromes, while 10 is not.
Flipping all bits in a substring means changing each 0 to 1 and each 1 to 0 in that substring. For example, flipping the substring 101 results in 010.
Input
Each test contains multiple test cases. The first line contains the number of test cases $T$ ($1 \le T \le 5\cdot 10^3$). The description of the test cases follows.
The first line of each test case contains an integer $n$ ($4 \le n \le 100$) — the length of the strings $s$ and $t$.
The next two lines of each test case contain the binary strings $s$ and $t$, respectively.
It is guaranteed that the sum of $n^2$ over all test cases does not exceed $5\cdot 10^5$.
Output
For each test case, if it is impossible to achieve the goal, print $-1$.
Otherwise, the first line should contain an integer $k$ ($0 \leq k \leq 2n$) — the number of operations.
For each of the next $k$ lines, print two integers $l, r$ ($1 \leq l \lt r \leq n$) — the indices you choose in each operation. Note that $s_{l,r}$ must be a palindrome at this stage.
Note
For the first test case:
- Initially,
$s = \mathtt{01011}$and$t = \mathtt{10000}$. - First, we choose
$l=1$and$r=3$. This is a valid operation since$s_{1,3} = \mathtt{010}$is a palindrome. After flipping$s_{1,3}$,$s = \mathtt{10111}$. - Then, we choose
$l=3$and$r=5$. This is a valid operation since$s_{3,5} = \mathtt{111}$is a palindrome. After flipping$s_{3,5}$,$s = \mathtt{10000}$. - Now,
$s$is equal to$t$.
For the second test case:
- Initially,
$s = \mathtt{1010101}$and$t = \mathtt{0101010}$. - First, we choose
$l=1$and$r=7$. This is a valid operation since$s_{1,7} = \mathtt{1010101}$is a palindrome. After flipping$s_{1,7}$,$s = \mathtt{0101010}$. - Now,
$s$is equal to$t$.
For the third test case:
- Initially,
$s$is equal to$t$. No operations are needed.
样例
3
5
01011
10000
7
1010101
0101010
4
0010
0010
2
1 3
3 5
1
1 7
0
题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。