You are given an array a[0 … n - 1] = [a_0, a_1, …, a_{n - 1}] of zeroes and ones only. Note that in this problem, unlike the others, the array indexes are numbered from zero, not from one.

In one step, the array a is replaced by another array of length n according to the following rules: 

  1. First, a new array a^{→ d} is defined as a cyclic shift of the array a to the right by d cells. The elements of this array can be defined as a^{→ d}_i = a_{(i + n - d) mod n}, where (i + n - d) mod n is the remainder of integer division of i + n - d by n. 

It means that the whole array a^{→ d} can be represented as a sequence $$$a^{→ d} = [a_{n - d}, a_{n - d + 1}, …, a_{n - 1}, a_0, a_1, …, a_{n - d - 1}]$$$

  2. Then each element of the array a_i is replaced by a_i  \&  a^{→ d}_i, where \& is a logical "AND" operator. 



For example, if a = [0, 0, 1, 1] and d = 1, then a^{→ d} = [1, 0, 0, 1] and the value of a after the first step will be [0  \&  1, 0  \&  0, 1  \&  0, 1  \&  1], that is [0, 0, 0, 1].

The process ends when the array stops changing. For a given array a, determine whether it will consist of only zeros at the end of the process. If yes, also find the number of steps the process will take before it finishes.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases.

The next 2t lines contain descriptions of the test cases. 

The first line of each test case description contains two integers: n (1 ≤ n ≤ 10^6) — array size and d (1 ≤ d ≤ n) — cyclic shift offset. The second line of the description contains n space-separated integers a_i (0 ≤ a_i ≤ 1) — elements of the array.

It is guaranteed that the sum of n over all test cases does not exceed 10^6.

Output

Print t lines, each line containing the answer to the corresponding test case. The answer to a test case should be a single integer — the number of steps after which the array will contain only zeros for the first time. If there are still elements equal to 1 in the array after the end of the process, print -1.

Example

Input


5
2 1
0 1
3 2
0 1 0
5 2
1 1 0 1 0
4 2
0 1 0 1
1 1
0


Output


1
1
3
-1
0

Note

In the third sample test case the array will change as follows: 

  1. At the beginning a = [1, 1, 0, 1, 0], and a^{→ 2} = [1, 0, 1, 1, 0]. Their element-by-element "AND" is equal to $$$[1  \&  1, 1  \&  0, 0  \&  1, 1  \&  1, 0  \&  0] = [1, 0, 0, 1, 0]$$$ 
  2. Now a = [1, 0, 0, 1, 0], then a^{→ 2} = [1, 0, 1, 0, 0]. Their element-by-element "AND" equals to $$$[1  \&  1, 0  \&  0, 0  \&  1, 1  \&  0, 0  \&  0] = [1, 0, 0, 0, 0]$$$ 
  3. And finally, when a = [1, 0, 0, 0, 0] we get a^{→ 2} = [0, 0, 1, 0, 0]. Their element-by-element "AND" equals to $$$[1  \&  0, 0  \&  0, 0  \&  1, 0  \&  0, 0  \&  0] = [0, 0, 0, 0, 0]$$$ 

Thus, the answer is 3 steps.

In the fourth sample test case, the array will not change as it shifts by 2 to the right, so each element will be calculated as 0  \&  0 or 1  \&  1 thus not changing its value. So the answer is -1, the array will never contain only zeros.
try:
        import sys
        from functools import lru_cache, cmp_to_key
        from heapq import merge, heapify, heappop, heappush
        # from math import *
        from collections import defaultdict as dd, deque, Counter as C
        from itertools import combinations as comb, permutations as perm
        from bisect import bisect_left as bl, bisect_right as br, bisect, insort
        from time import perf_counter
        from fractions import Fraction
        import copy
        from copy import deepcopy
        import time
        starttime = time.time()
        mod = int(pow(10, 9) + 7)
        mod2 = 998244353

        def data(): return sys.stdin.readline().strip()
        def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
        def L(): return list(sp())
        def sl(): return list(ssp())
        def sp(): return map(int, data().split())
        def ssp(): return map(str, data().split())
        def l1d(n, val=0): return [val for i in range(n)]
        def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
        def A2(n,m): return [[0]*m for i in range(n)]
        def A(n):return [0]*n
        # input = stdin.buffer.readline
        # I = lambda : list(map(int,input().split()))
        # import sys
        # input=sys.stdin.readline
        

        sys.stdin = open("input.txt", "r")
        sys.stdout = open("output.txt", "w")
except:
        pass



for _ in range(L()[0]):
        n,d = L()
        A = L()

        dist =  [-A[i] for i in range(n)]
        for i in range(n):
                if dist[(i-d)%n]!=-1 and dist[i]==-1:
                        dist[i] = dist[(i-d)%n] + 1
        for i in range(n):
                if dist[(i-d)%n]!=-1 and dist[i]==-1:
                        dist[i] = dist[(i-d)%n] + 1
                        
        if -1 in dist:
                print(-1)
        else:
                print(max(dist))