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 sys,os,io
from sys import stdin
if(os.path.exists('input.txt')):
    sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w") 
else:
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

t = int(input())


for _ in range(t):
    n,k = [int(x) for x in input().split()]
    a = [int(x) for x in input().split()]
    dp = [[0 for i in range(n+2)] for j in range(n+2)]
    for i in range(0,n+1):
        for j in range(i+1):
            dif = a[i-1]
            dp[i+1][j]=max(dp[i+1][j],dp[i][j])
            temp = 0
            if dif==j:
                temp = 1
            dp[i+1][j+1]=max(dp[i+1][j+1],dp[i][j]+temp)
            # dp[i][j]=max(dp[i][j],dp[i-1][j],dp[i-1][j-1])
            # if dif==j:
            #     if i>0 and j>0:
            #         dp[i][j]=max(1+dp[i-1][j-1],dp[i][j])
            #     else:
            #         dp[i][j]=max(1,dp[i][j])
    res = float('inf')
    # for i in dp:
    #     print(*i)
    for i in range(n+1):
        if dp[-1][i]>=k:
            res=min(res,n-i)
            break 
    if res==float('inf'):
        res = -1
    print(res)
    # pos = [[0 for i in range(n+1)] for j in range(n+1)]
    # res = 10000000000000000000
    # def rec(i,j,a,k):
    #     global res
    #     #till index i , j elements are removed
    #     if i==len(a):
    #         return 0
    #     diff = -(a[i]-(i+1))
    #     # print(i,j,diff)
    #     if diff!=j:
    #         ans = max(rec(i+1,j+1,a,k),rec(i+1,j,a,k))
    #         if ans>=k:
    #             res = min(res,j)    
    #         return ans
    #     ans = max(rec(i+1,j+1,a,k),rec(i+1,j,a,k))
    #     if diff == j:
    #         ans = max(ans, max(rec(i+1,j+1,a,k),rec(i+1,j,a,k))+1)
    #     if ans>=k:
    #         pos[i][j]=1
    #     return ans
    # rec(0,0,a,k)
    # print(pos)
    # ans = 1000000000000000
    # for i in range(n+1):
    #     if pos[-2][i]==1:
    #         ans = min(ans,i)
    # print(ans)