Consider a sequence of integers a_1, a_2, …, a_n. In one move, you can select any element of the sequence and delete it. After an element is deleted, all elements to the right are shifted to the left by 1 position, so there are no empty spaces in the sequence. So after you make a move, the sequence's length decreases by 1. The indices of the elements after the move are recalculated.

E. g. let the sequence be a=[3, 2, 2, 1, 5]. Let's select the element a_3=2 in a move. Then after the move the sequence will be equal to a=[3, 2, 1, 5], so the 3-rd element of the new sequence will be a_3=1 and the 4-th element will be a_4=5.

You are given a sequence a_1, a_2, …, a_n and a number k. You need to find the minimum number of moves you have to make so that in the resulting sequence there will be at least k elements that are equal to their indices, i. e. the resulting sequence b_1, b_2, …, b_m will contain at least k indices i such that b_i = i.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

Each test case consists of two consecutive lines. The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2000). The second line contains a sequence of integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). The numbers in the sequence are not necessarily different.

It is guaranteed that the sum of n over all test cases doesn't exceed 2000.

Output

For each test case output in a single line:

  * -1 if there's no desired move sequence; 
  * otherwise, the integer x (0 ≤ x ≤ n) — the minimum number of the moves to be made so that the resulting sequence will contain at least k elements that are equal to their indices. 

Example

Input


4
7 6
1 1 2 3 4 5 6
5 2
5 1 3 2 3
5 2
5 5 5 5 4
8 4
1 2 3 3 2 2 5 5


Output


1
2
-1
2

Note

In the first test case the sequence doesn't satisfy the desired condition, but it can be provided by deleting the first element, hence the sequence will be [1, 2, 3, 4, 5, 6] and 6 elements will be equal to their indices.

In the second test case there are two ways to get the desired result in 2 moves: the first one is to delete the 1-st and the 3-rd elements so that the sequence will be [1, 2, 3] and have 3 elements equal to their indices; the second way is to delete the 2-nd and the 3-rd elements to get the sequence [5, 2, 3] with 2 desired elements.
import io,os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline


def judge(remove,arr,k):

    n = len(arr)
    
    dp = [[0 for j in range(remove+1)] for i in range(n+1)]
    for i in range(n-1,-1,-1):
        limit = min(i,remove)
        for j in range( limit+1 ):
             
            if arr[i] ==  j:  base = 1
            else: base = 0
 
            num1 = base + dp[i+1][j]
            num2 = -2147483647
            if j+1<=remove:  num2 = dp[i+1][j+1]
            
            dp[i][j] = max(num1,num2)

#    print(remove)
#    print(dp[0][0])
#    print(dp)
#    print("-")

    if dp[0][0] >= k: return True
    else: return False



            
            






def main(t):


    n,k = map(int,input().split())
    arr = list(map(int,input().split()))

    for i in range(n):
        arr[i] = i+1 - arr[i]

#    print(arr)


    ans = 0
    front = 0
    rear = n-k+1
    while front<rear:
        mid = (front+rear)//2
     
        if judge(mid,arr,k):  rear = mid 
        else:  front = mid + 1




    if front==n-k+1: print(-1)
    else: print(front)























T = int(input())
t = 1
while t<=T:
    main(t)
    t += 1
