You can check out the explanation by ninjacoder2209 in this video, he has explained it very nicely as well. It might give you some more clearity

https://www.youtube.com/watch?v=2M5S9QNetg4

We can derive a formula for each of the points in the array a to find out in how many segments they exist. Now for the numbers between each element of a (if any exist) the formula for how many segments they exist in is slightly different.

Anyhow, once we have the said formula, we can start looping through the array a and in a map store the number of segments the element ai exists in and increment it if any other ai exists the same number of times. For the numbers between ai, we do the same thing.

Once we have this map of segments and the count of numbers that exist in these segments we can loop through queries, see if they exist and then output the answer accordingly.

Here's a python code that breaks it down and I have added comments as well that break down the formula derivation.

t = int(input())
for i in range(t):
    n, q = list(map(int, input().split(" ")))
    a = list(map(int, input().split(" ")))
    k = list(map(int, input().split(" ")))

    ans = {}

    l = n
    n -= 1
    for i in range(l-1):
        ans[a[i]] = n-i + i*(n-i+1)

        if a[i+1] - a[i] > 1:
            ans[a[i]+1] = n-i + i*(n-i)
    ans[a[i+1]] = n

    print(ans)

    """
    1 2 3 5 6 7

    1 = 5
    2 = 4 + 5                   = 9
    3 = 3 + 4 + 4 or (5-1)      = 11
    # 4 = 3 + 3 or (4-1) + 3 or (4-1) = 9
    5 = 2 + 3 + 3 + 3 = 11
    7 = 1 + 1 + 1 + 1 + 1 =

    1 = n
    2 = n-1 + n
    3 = n-2 + 2*(n-1)
    .
    4 = n-2 + 2*(n-2)
    .
    5 = n-3 + 3*(n-2)
    6 = n-4 + 4*(n-3)
    7 = n-5 + 5*(n-4)
    """