You are given $N$ integers in an array: $A[1], A[2], \ldots, A[N]$. You also have another integer $L$.
Consider a sequence of indices ($i_1, i_2, \ldots, i_k$). Note that a particular index can occur multiple times in the sequence, and there is no order in which these indices have to occur. ($i_1, i_2, \ldots, i_k$) is a sequence of size $k$. It is said to be an $Interesting$ sequence, if $A[i_1] \ge A[i_2] \ge \ldots \ge A[i_k]$.
The $Cost$ of an Interesting sequence ($i_1, i_2, \ldots, i_k$), is defined to be the minimum absolute difference between any two adjacent indices. In other words, the Cost is $min \{ |i_2 - i_1|, |i_3 - i_2|, \ldots, |i_k - i_{k-1}| \}$.
Your job is to consider the Costs of all the Interesting sequences of size $L$ associated with the given array, and output the maximum Cost. Note that you can show that there is always at least one Interesting sequence for the given constraints.
-----Input-----
- The first line contains a single integer, $T$, which is the number of testcases. The description of each testcase follows.
- The first line of each testcase contains two space separated integers: $N$ and $L$.
- The second line of each testcase contains $N$ space separated integers: $A[1], A[2], \ldots, A[N]$.
-----Output-----
- For each testcase, output the answer in a new line.
-----Constraints-----
- $1 \leq T \leq 3$
- $1 \leq A[i] \leq 10^9$
- $2 \leq L \leq 10^9$
-----Subtasks-----
- Subtask 1: 7 points
- It is guaranteed that $A[1] > A[2] > \ldots > A[N]$
- Note that the above condition implies that all elements are distinct.
- $1 \leq N \leq 500$
- Subtask 2: 7 points
- It is guaranteed that $A[1] \ge A[2] \ge \ldots \ge A[N]$
- $1 \leq N \leq 500$
- Subtask 3: 14 points
- It is guaranteed that all elements are distinct.
- $1 \leq N \leq 500$
- Subtask 4: 14 points
- $1 \leq N \leq 500$
- Subtask 5: 25 points
- It is guaranteed that all elements are distinct.
- $1 \leq N \leq 3000$
- Subtask 6: 33 points
- $1 \leq N \leq 3000$
-----Sample Input-----
1
6 3
2 4 1 12 3 5
-----Sample Output-----
3
-----Explanation-----
We are looking for Interesting sequences of length 3. Some of them are:
- (4, 2, 3): This is Interesting because $A[4] \ge A[2] \ge A[3]$. Its cost is $min \{ |2-4|, |3-2|\} = 1$.
- (5, 1, 1): Cost is 0.
- (2, 2, 2): Cost is 0.
- (6, 1, 3): Cost is 2.
- (6, 2, 5): Cost is 3.
There are other Interesting Sequences of length 3 as well. But if you list them all out, you'll see that the maximum Cost is 3. Hence the answer is 3.
# cook your dish here
# from queue import PriorityQueue
import bisect
for _ in range(int(input())):
    n,l = map(int, input().split())
    a_l = list(map(int, input().split()))
    
    dic = {}
    dif = 0
    for i,v in enumerate(a_l, start=1):
        if v not in dic:
            dic[v] = [i, i]
        else:
            dic[v][0] = min(dic[v][0], i)
            dic[v][1] = max(dic[v][1], i)
            dif = max(dif, dic[v][1]-dic[v][0])
            
    ans = dif
    if l <= len(set(a_l)):
        i_l = [[v,i] for i,v in enumerate(a_l, start=1)]
        i_l.sort(reverse=True)
        
        dp = [[-1 for _ in range(l)] for _ in range(n)]
        pq_l = [[] for _ in range(l)]
        for i in range(1,n):
            il = 1
            dif_l = []
            for j in range(i):
                dif = abs(i_l[i][1]-i_l[j][1])
                dif_l.append(dif)
                dp[i][il] = max(dp[i][il], dif)
                
            for il in range(2,min(l,i+1)):
                for prev_max, ind in reversed(pq_l[il-1]):
                    if ind == i:
                        continue
                    if prev_max < dp[i][il]:
                        break
                    else:
                        dp[i][il] = max(dp[i][il], min(dif_l[ind], prev_max))
                tmp = [v[0] for v in pq_l[il]]
                ind = bisect.bisect_right(tmp, dp[i][il])
                pq_l[il].insert(ind, [dp[i][il], i])
                
            il = 1
            tmp = [v[0] for v in pq_l[il]]
            ind = bisect.bisect_right(tmp, dp[i][il])
            pq_l[il].insert(ind, [dp[i][il], i])
            
            # print(i, pq_l, dif_l)
                        
                
                # dp[i][1] = max(dp[i][1], dif)
                # for il in range(2,l):
                #     if dp[j][il-1] == -1:
                #         break
                #     dp[i][il] = max(dp[i][il], min(dif, dp[j][il-1]))
            ans = max(ans, dp[i][-1])
    #     print(dp)
    # print(dic)
    print(ans)
# cook your dish here
# from queue import PriorityQueue
import bisect
for _ in range(int(input())):
    n,l = map(int, input().split())
    a_l = list(map(int, input().split()))
    
    dic = {}
    dif = 0
    for i,v in enumerate(a_l, start=1):
        if v not in dic:
            dic[v] = [i, i]
        else:
            dic[v][0] = min(dic[v][0], i)
            dic[v][1] = max(dic[v][1], i)
            dif = max(dif, dic[v][1]-dic[v][0])
            
    ans = dif
    if l <= len(set(a_l)):
        i_l = [[v,i] for i,v in enumerate(a_l, start=1)]
        i_l.sort(reverse=True)
        
        dp = [[-1 for _ in range(l)] for _ in range(n)]
        pq_l = [[] for _ in range(l)]
        for i in range(1,n):
            il = 1
            dif_l = []
            for j in range(i):
                dif = abs(i_l[i][1]-i_l[j][1])
                dif_l.append(dif)
                dp[i][il] = max(dp[i][il], dif)
                
            for il in range(2,min(l,i+1)):
                for prev_max, ind in reversed(pq_l[il-1]):
                    if ind == i:
                        continue
                    if prev_max < dp[i][il]:
                        break
                    else:
                        dp[i][il] = max(dp[i][il], min(dif_l[ind], prev_max))
                tmp = [v[0] for v in pq_l[il]]
                ind = bisect.bisect_right(tmp, dp[i][il])
                pq_l[il].insert(ind, [dp[i][il], i])
                
            il = 1
            tmp = [v[0] for v in pq_l[il]]
            ind = bisect.bisect_right(tmp, dp[i][il])
            pq_l[il].insert(ind, [dp[i][il], i])
            
            # print(i, pq_l, dif_l)
                        
                
                # dp[i][1] = max(dp[i][1], dif)
                # for il in range(2,l):
                #     if dp[j][il-1] == -1:
                #         break
                #     dp[i][il] = max(dp[i][il], min(dif, dp[j][il-1]))
            ans = max(ans, dp[i][-1])
    #     print(dp)
    # print(dic)
    print(ans)