You are given a non-decreasing array of non-negative integers a_1, a_2, …, a_n. Also you are given a positive integer k.

You want to find m non-decreasing arrays of non-negative integers b_1, b_2, …, b_m, such that:

  * The size of b_i is equal to n for all 1 ≤ i ≤ m. 
  * For all 1 ≤ j ≤ n, a_j = b_{1, j} + b_{2, j} + … + b_{m, j}. In the other word, array a is the sum of arrays b_i. 
  * The number of different elements in the array b_i is at most k for all 1 ≤ i ≤ m. 



Find the minimum possible value of m, or report that there is no possible m.

Input

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

The first line of each test case contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ n).

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_1 ≤ a_2 ≤ … ≤ a_n ≤ 100, a_n > 0).

Output

For each test case print a single integer: the minimum possible value of m. If there is no such m, print -1.

Example

Input


6
4 1
0 0 0 1
3 1
3 3 3
11 3
0 1 2 2 3 3 3 4 4 4 4
5 3
1 2 3 4 5
9 4
2 2 3 5 7 11 13 13 17
10 7
0 1 1 2 3 3 4 5 5 6


Output


-1
1
2
2
2
1

Note

In the first test case, there is no possible m, because all elements of all arrays should be equal to 0. But in this case, it is impossible to get a_4 = 1 as the sum of zeros.

In the second test case, we can take b_1 = [3, 3, 3]. 1 is the smallest possible value of m.

In the third test case, we can take b_1 = [0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2] and b_2 = [0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2]. It's easy to see, that a_i = b_{1, i} + b_{2, i} for all i and the number of different elements in b_1 and in b_2 is equal to 3 (so it is at most 3). It can be proven that 2 is the smallest possible value of m.
#   Author: yumtam
#   Created at: 2020-10-15 10:10

from __future__ import division, print_function
_interactive = False

def main():
    def ceil_div(a, b):
        return a // b + (a % b > 0)

    for _ in range(int(input())):
        n, k = input_as_list()
        ar = input_as_list()

        dif = len(set(ar))

        if k == 1:
            ans = 1 if dif == 1 else -1
        else:
            ans = max(ceil_div(dif-1, k-1), 1)
        print(ans)


# Constants
INF = float('inf')
MOD = 10**9+7

# Python3 equivalent names
import os, sys, itertools
if sys.version_info[0] < 3:
    input = raw_input
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip

# print-flush in interactive problems
if _interactive:
    flush = sys.stdout.flush
    def printf(*args, **kwargs):
        print(*args, **kwargs)
        flush()

# Debug print, only works on local machine
LOCAL = "LOCAL_" in os.environ
debug_print = (print) if LOCAL else (lambda *x, **y: None)

# Fast IO
if (not LOCAL) and (not _interactive):
    from io import BytesIO
    from atexit import register
    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
    sys.stdout = BytesIO()
    register(lambda: os.write(1, sys.stdout.getvalue()))
    input = lambda: sys.stdin.readline().rstrip('\r\n')

# Some utility functions(Input, N-dimensional lists, ...)
def input_as_list():
    return [int(x) for x in input().split()]

def input_with_offset(o):
    return [int(x)+o for x in input().split()]

def input_as_matrix(n, m):
    return [input_as_list() for _ in range(n)]

def array_of(f, *dim):
    return [array_of(f, *dim[1:]) for _ in range(dim[0])] if dim else f()

# Start of external code templates...
# End of external code templates.

main()
