human
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...
The above is tricky. Write me a correct solution in PYTHON3.
 ... Give me a 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?
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.

from collections import deque
import heapq
from math import sqrt
import operator
from operator import itemgetter
import sys

import copy

 
inf_var = 0
if inf_var == 1:
    inf = open('input.txt', 'r')
else:
    inf = sys.stdin
 
input = inf.readline
# sys.stdout = open('output.txt', 'w')
 
def read_one_int():
    return int(input().rstrip('\n'))

def read_one_float():
    return float(input().rstrip('\n'))
 
 
def read_list_of_ints():
    res = [int(val) for val in (input().rstrip('\n')).split(' ')]
    return res
 
def read_str():
    return input().rstrip()

def main():
    tests = read_one_int()

    for _ in range(tests):
        n, k = read_list_of_ints()
        elements = read_list_of_ints()

        graph = [[] for _ in range(n)]

        for i in range(n - 1, -1, -1):
            need_to_del_i = i + 1 - elements[i]
            if need_to_del_i < 0:
                continue
            for j in range(i - 1, -1, -1):
                need_to_del_j = j + 1 - elements[j]

                if need_to_del_j < 0 or need_to_del_j > need_to_del_i:
                    continue

                if j < need_to_del_j or need_to_del_i - need_to_del_j > i - j - 1:
                    continue

                graph[i].append(j)

        # print(graph)
        cache = {}

        def count_max(v):
            if v in cache:
                return cache[v]

            if v + 1 - elements[v] >= 0:
                res = 0
                for u in graph[v]:
                    res = max(res, count_max(u))

                cache[v] = res + 1
                return res + 1
            else:
                cache[v] = 0
                return 0            

        res = -1
        for i in range(n):
            cur_res = count_max(i)

            if cur_res >= k:
                if res == -1:
                    res = i + 1 - elements[i]
                else:
                    res = min(res, i + 1 - elements[i])

        print(res)

if __name__== '__main__':
    main()