After six days, professor GukiZ decided to give more candies to his students. Like last time, he has $N$ students, numbered $1$ through $N$. Let's denote the number of candies GukiZ gave to the $i$-th student by $p_i$. As GukiZ has a lot of students, he does not remember all the exact numbers of candies he gave to the students. He only remembers the following properties of the sequence $p$:
- The numbers of candies given to each of the first $K$ students ($p_1, p_2, \dots, p_K$) are known exactly.
- All elements of the sequence $p$ are distinct and positive.
- GukiZ didn't give more than $x$ candies to any student (the maximum value in the sequence $p$ is not greater than $x$).
- For each student $i$, there is at least one other student $j$ such that $|p_i - p_j| \le D$.
- The professor gave out the biggest possible total number of candies, i.e. $S = p_1 + p_2 + p_3 + \ldots + p_N$ is maximum possible.
GukiZ would like to know the total number of candies $S$ he had at the beginning. However, times change and after six days, the professor is really tired, so it is possible that there is no sequence $p$ which satisfies the constraints. Can you help GukiZ find the number of candies he gave out, or tell him that he must have made a mistake?
-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first line of each test case contains four space-separated integers $N$, $K$, $x$, $D$.
- The second line contains $K$ distinct space-separated integers $p_1, p_2, \dots, p_K$.
-----Output-----
For each test case, print a single line containing one integer — the number of candies GukiZ had, or $-1$ if there is no valid sequence $p$.
-----Constraints-----
- $1 \le T \le 50$
- $3 \le N \le 10^9$
- $1 \le K \le \mathrm{min}(N, 2 \cdot 10^4)$ 
- $1 \le x \le 2 \cdot 10^9$
- $1 \le D \le 10^9$
- $1 \le p_i \le x$ for each valid $i$
- All values $p_i$ from input are distinct
-----Subtasks-----
Subtask #1 (15 points): $1 \leq x, N, D \leq 15$ 
Subtask #2 (35 points): $1 \leq x, N, D \leq 10^5$ 
Subtask #3 (50 points): original constraints
-----Example Input-----
2
4 3 5 3
2 1 5
3 2 8 2
3 8
-----Example Output-----
12
-1
-----Explanation-----
Example case 1: There are four students. We know that the first student got $p_1 = 2$ candies, the second student got $p_2 = 1$ and the third got $p_3 = 5$ candies; we don't know the number of candies given to the last student. The maximum possible amount of candies given to some student is $x=5$. The best possible option is giving $p_4=4$ candies to the last student. Then, the fourth constraint (with $D=3$) is satisfied for all students. Only the pair of students $(2, 3)$ have numbers of candies that differ by more than $3$, but still, for each student, there are at least two other students with close enough numbers of candies.
Example case 2: GukiZ made some mistake in distribution and there is no valid sequence $p$. The answer is $-1$.
# https://www.codechef.com/LTIME63B/problems/GHMC
# Finally.... I properly understood what needs to be done.

def ctlt(arr, val):
    # find number of values in sorted arr < val
    if arr[0] >= val: return 0
    lo = 0
    hi = len(arr)
    while hi-lo > 1:
        md = (hi+lo)//2
        if arr[md]<val:
            lo = md
        else:
            hi = md

    return hi

for _ in range(int(input())):
    n,k,x,d = map(int, input().split())
    z = input().strip().split()
    if k > 0:
        ps = list(map(int,z[:k]))
    else:
        ps = [x]

    ps.sort()

    if x<n or x<ps[-1] or n<k:
        print(-1)
        continue

    valchecked = 0
    fillval = 0
    valsdone = False
    isolbelow = True
    lastp = ps[0]

    while not valsdone and n>=k:
        if n == k:
            lo = x+d+1 # put out of range
        else:
            # find best maxfill (before val support)
            lo = 1
            hi = x+1
            while hi-lo>1:
                md = (hi+lo)//2
                v = (x-md+1) + ctlt(ps,md)
                if v<n:
                    hi = md
                else:
                    lo = md
        
        valsdone = True
        checkto = ctlt(ps,lo)-1
        if checkto >= valchecked:            
            # support all vals
            for p in ps[valchecked+1:checkto+1]:
                if lastp+d >= p:
                    isolbelow = False
                elif isolbelow:
                    valsdone = False
                    fillval += lastp+d
                    n -= 1
                    isolbelow = (p > lastp + 2*d )
                else:
                    isolbelow = True
                lastp = p
            valchecked = checkto
            if valsdone and isolbelow: 
                # check gap to maxfill
                if lastp + d >= lo:
                    isolbelow = False
                else:
                    valsdone = False
                    fillval += lastp
                    ps[checkto] += d
                    lastp += d
                    isolbelow = False
                    n -= 1

    if k > n:
        print(-1)
    elif k == n:
        print(sum(ps) + fillval)
    elif k == n-1 and lo > ps[-1]:
        print(sum(ps) + fillval + min(x,ps[-1]+d))
    else:
        tot = (x+lo)*(x-lo+1)//2  + sum(ps[:ctlt(ps,lo)])
        print(tot + fillval)

# https://www.codechef.com/LTIME63B/problems/GHMC
# Finally.... I properly understood what needs to be done.

def ctlt(arr, val):
    # find number of values in sorted arr < val
    if arr[0] >= val: return 0
    lo = 0
    hi = len(arr)
    while hi-lo > 1:
        md = (hi+lo)//2
        if arr[md]<val:
            lo = md
        else:
            hi = md

    return hi

for _ in range(int(input())):
    n,k,x,d = map(int, input().split())
    z = input().strip().split()
    if k > 0:
        ps = list(map(int,z[:k]))
    else:
        ps = [x]

    ps.sort()

    if x<n or x<ps[-1] or n<k:
        print(-1)
        continue

    valchecked = 0
    fillval = 0
    valsdone = False
    isolbelow = True
    lastp = ps[0]

    while not valsdone and n>=k:
        if n == k:
            lo = x+d+1 # put out of range
        else:
            # find best maxfill (before val support)
            lo = 1
            hi = x+1
            while hi-lo>1:
                md = (hi+lo)//2
                v = (x-md+1) + ctlt(ps,md)
                if v<n:
                    hi = md
                else:
                    lo = md
        
        valsdone = True
        checkto = ctlt(ps,lo)-1
        if checkto >= valchecked:            
            # support all vals
            for p in ps[valchecked+1:checkto+1]:
                if lastp+d >= p:
                    isolbelow = False
                elif isolbelow:
                    valsdone = False
                    fillval += lastp+d
                    n -= 1
                    isolbelow = (p > lastp + 2*d )
                else:
                    isolbelow = True
                lastp = p
            valchecked = checkto
            if valsdone and isolbelow: 
                # check gap to maxfill
                if lastp + d >= lo:
                    isolbelow = False
                else:
                    valsdone = False
                    fillval += lastp
                    ps[checkto] += d
                    lastp += d
                    isolbelow = False
                    n -= 1

    if k > n:
        print(-1)
    elif k == n:
        print(sum(ps) + fillval)
    elif k == n-1 and lo > ps[-1]:
        print(sum(ps) + fillval + min(x,ps[-1]+d))
    else:
        tot = (x+lo)*(x-lo+1)//2  + sum(ps[:ctlt(ps,lo)])
        print(tot + fillval)
