Problem
You are given two non-decreasing sequences A = (A₁, A2,..., AM) and B = (B₁, B2,..., BN).
You can choose any two indices i and j, and then swap A[i]; and B[j].
Note that the you can do the operations as many times as you want, and your goal is to transform A into an
arithmetic sequence. After opeartions, you can rearrange the sequence A.
Now, determine how many distinct arithmetic sequences you can obtain.
Input Format
• The first line contains a single integer T - the number of test cases.
• For each testcase:
• The first line contains two integer M and N-the lengths of the sequences A and B, respectively.
• The second line contains M integers A₁, A2,..., AM.
• The third line contains N integers B₁, B2,..., BN.
Output Format
• For each testcase, output the number of distinct arithmetic sequences you can obtain.
Constraints
• 1 ≤ T ≤ 10³
• 2 ≤ N ≤ M ≤ 105
• -M ≤ A₁, B₁ ≤ M
• The sum of M + N across all test cases does not exceed 2. 10^7.
sample input
33
-10 2
0 1 2
33
0 0 1
0 1 1
Sample Output
4
2
8
out put is : 4

observation

Because you can swap A[i] and B[j] as many times as you want, your goal is to form distinct arithmetic sequences of length M.

Steps

First, since you are given two sorted sequences, you can merge these two into one in O(M).

Now, you have a sequence C of length M + N, and consider two kinds of arithmetic sequences:

• Common difference is = 0:

That is, the sequence is written as (a, a, ..., a). To find this kind of sequences, count each element in C.

If there are at least M's a in C, then add the desired answer by 1.

• Common difference is != 0:

Also, there may be some arithmetic sequences of the form (a, a + d, a + 2 · d, ..., a + (M − 1) · d) for some d != 0.

In this case, elements in one arithmetic sequence are all different, so please remove duplicate elements in C.

Since the sequence C is non-decreasing, consider the case d > 0 here.

Moreover, since - M ≤ Ai, B; ≤ M and the arithmetic sequence should be of length M, the choices of the common difference d > 0 is finite. In fact, the choices is at

most

2.M/M-1

So, for each possible common difference d, please count the number of arithmetic sequences.

Note that for each common difference d > 0, there would be the corresponding negative common difference -d.

Thus, please multiple the number by 2 and then add this number into the answer.

so if the a =[-1,0,1] and b=[0,1,2] and m = 3 I want to know according to give formula 2.m/m-1 (because the merge sequence of this common difference is != 0). which give no of possible choices. how can I obtain the d values. to create the arithmetic sequences.?

**What I have tried:**
*so if a =[-1,0,1] and b=[0,1,2] and m = 3 according to give formula 2.m/m-1 (because the merge sequence of this common difference is != 0). which give number of possible choices. i get the value 3 so there should be 3 possible choices.*
here is where I am at so I merged the 2 none decreasing sequences to get the sequence length M+N which C = [-1,0,1,0,1,2] cause the common difference is != 0 I went with the 2nd method("• Common difference is != 0:").

I removed the duplicands and sorted the sequence. so now I got c =[-1,0,1,2] so the sequence values a[1] = -1 , -m <= a[i] if m= 3 I understand -3<=-1 and b[j]<= m which b[1] =1 and 1<= 3. according to the solution and it says that sequence it continue to m. since its none decreasing we consider values d > 0 only.

so it say to get the number of the possible "d" values to create the distinct sequences use the formula 2.M/M-1 and I got answer 3, 3 possible choices form here i only know that the output is [-1,0,1] and [0,1,2] and there negative which counts the total number of 4 distinct arithmetic sequences. I don't know how to get the d values to create the above sequence.

(assumption :- if the possible choice is 3 should i just use 1,2,3 and add each element in to the first of the sequence -1 and continue -1,0,1 or something else? this is where i'm stuck and with out this I can't code)