human
PYTHON3 solution for "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...":
assistant
#Here is Python code 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.
 def var():
    return (map(int,input().split()))
def a():
    return list((map(int,input().split())))
def inp():
    return int(input())

import math
import copy
from collections import defaultdict,deque
import heapq
import sys
# input = sys.stdin.readline
sys.setrecursionlimit(10**5)
mod = 10**9+7
input = sys.stdin.buffer.readline

t=inp()
for _ in range(t):
    n,k=var()
    arr=a()
    
    dp=[[0]*(n+1) for _ in range(n+1)]
    dp[0][0]=0
    for i in range(1,n+1):
        for j in range(i+1):
            if j==0:
                dp[i][j]=dp[i-1][j]+int((arr[i-1])==i)
                continue
            dp[i][j]=max(dp[i-1][j-1],dp[i-1][j]+int(arr[i-1]==(i-j)))
    
    found=False
    for j in range(n+1):
        if dp[n][j]>=k:
            found=True
            break
    if found:
        print(j)
    else:
        print(-1)