Shubham has an array a of size n, and wants to select exactly x elements from it, such that their sum is odd. These elements do not have to be consecutive. The elements of the array are not guaranteed to be distinct.

Tell him whether he can do so.

Input

The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers n and x (1 ≤ x ≤ n ≤ 1000) — the length of the array and the number of elements you need to choose.

The next line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — elements of the array.

Output

For each test case, print "Yes" or "No" depending on whether it is possible to choose x elements such that their sum is odd.

You may print every letter in any case you want.

Example

Input


5
1 1
999
1 1
1000
2 1
51 50
2 2
51 50
3 3
101 102 103


Output


Yes
No
Yes
Yes
No

Note

For 1st case: We must select element 999, and the sum is odd.

For 2nd case: We must select element 1000, so overall sum is not odd.

For 3rd case: We can select element 51.

For 4th case: We must select both elements 50 and 51 — so overall sum is odd.

For 5th case: We must select all elements — but overall sum is not odd.

# answer_list = list()
# for times in range(int(input())):
#     requirements = input().split(' ')
#     array_a = [False for i in input().split(' ') if int(i)%2 != 0 ]
#     length = int(requirements[0])
#     pick = int(requirements[1])

#     odd = len(array_a)
#     even = length - odd

#     # you need at least 1 odd number
#     # then the rest of the odd number ,
#     # i can use 2 odd number transform into a even number
#     if odd == 0:
#         # print("No")
#         answer_list.append("No")
#     else:
#         #we pick one odd number already
#         pick = pick - 1
#         if even - pick >= 0:
#             answer_list.append('Yes')
#         else:
#             #if not enough , then transform 2 odd into 1 even
#             #but also remember that means you pick two numbers
#             odd  = odd-1 # as you already pick one earlier
#             #if there is only 1 pick left
#             # there is no way you can pick an even number which is transformed by two odd nubmers
#             # the int() transform float number by just dropping the decimal part
#             #print('pick : ',pick , 'even : ',even , 'odd : ', odd)
#             left_to_pick = int( (pick - even) / 2)
#             #print('left to pick : ',left_to_pick)
#             transformed_even_left = int(odd // 2)
#             #print('transformed_even_left : ',transformed_even_left)
#             if left_to_pick == 0:
#                 answer_list.append('No')

#             elif left_to_pick >= transformed_even_left:
#                 answer_list.append('Yes')
#             else:
#                 answer_list.append('No')




# for i in answer_list:
#     print(i)

#==========================================================================
#the above have some problems 



answer_list = list()
for times in range(int(input())):
    requirements = input().split(' ')
    array_a = [False for i in input().split(' ') if int(i)%2 != 0 ]
    length = int(requirements[0])
    pick = int(requirements[1])

    odd = len(array_a)
    even = length - odd

    # you need at least 1 odd number
    # then the rest of the odd number ,
    # i can use 2 odd number transform into a even number
    if odd == 0:
        # print("No")
        answer_list.append("No")
    else:
        #we pick one odd number already
        check = 0
        for odd_number in range(1,odd+1,2):
            
            left_to_pick = pick-odd_number

            if ( left_to_pick <= even )and (left_to_pick >= 0):
                # print("trigger Yes : ",odd_number,left_to_pick,even)
                answer_list.append("Yes")
                check = 1
                break

        if check != 1:
            answer_list.append("No")
        else:
            check = 0
for i in answer_list:
        print(i)
