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...":


 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.
 def solve(A, k):
    n = len(A)

    dp = [[0] * (n + 1) for _ in range(n + 1)]

    for i in range(n):
        for l in range(1, i + 2):
            dp[i][l] = max(dp[i][l], dp[i - 1][l])
                
            d = 1 if (A[i] == l) else 0
            dp[i][l] = max(dp[i][l], dp[i - 1][l - 1] + d)

    for i in reversed(range(n + 1)):
        if dp[n - 1][i] >= k:
            return n - i

    return -1

T = int(input())
for _ in range(T):
    n, k = map(int, input().split(' '))
    A = list(map(int, input().split(' ')))

    print(solve(A, k))