Problem: 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.

Solution (in PYTHON3):

Solution:
def main():
    #from collections import defaultdict
    #from collections import Counter
    #import math
    # from fractions import Fraction
    # import heapq
    #from bisect import bisect_left
    import sys
    input = sys.stdin.buffer.readline
    #mod=998244353
    def GCD(x, y):
        while (y):
            x, y = y, x % y
        return x
    def lcm(a,b):
        return (a*b)//GCD(a,b)
    # Modified merge for inversion
    def merge(a,start,mid,end,d):
        p=start
        q=mid+1
        arr=[0]*(end-start+1)
        k=0
        coun=0
        for i in range(start,end+1):
            if p > mid:
                arr[k]=a[q]
                q+=1
                k+=1
            elif q > end:
                arr[k]=a[p]
                d[a[p]] += coun
                k+=1
                p+=1
            elif a[p] < a[q]:
                arr[k]=a[p]
                d[a[p]]+=coun
                k+=1
                p+=1
            else:
                arr[k]=a[q]
                k+=1
                q+=1
                coun+=1

        for i in range(k):
            a[start]=arr[i]
            start+=1
    def mergesort(a, start, end,d):
        if start<end:
            mid=(start+end)//2
            mergesort(a,start,mid,d)
            mergesort(a,mid+1,end,d)
            merge(a,start,mid,end,d)
    # inversion return dic for every index inversion count
    def inversion(l):
        d={}
        for i in l:
            d[i]=0
        mergesort(l,0,len(l)-1,d)
        return d
    def binary_search_lower(A,low,high,k):
        if high-low!=1:
            mid=(low+high)//2
            x = A[mid]
            if x>=k:
                return binary_search_lower(A,low,mid,k)
            else:
                return binary_search_lower(A,mid,high,k)
        if A[low]>=k:
            return low
        else:
            return high
    # Function to find power
    def binary_expo(x,n):
        if n==0:
            return 1
        elif n%2==0:
            return binary_expo(x*x,n//2)
        else:
            return x*binary_expo(x*x,(n-1)//2)
    # Function to find power with given modulus
    def modu_expo(x,n,m):
        if n==0:
            return 1
        elif n%2==0:
            return modu_expo((x*x)%m,n//2,m)
        else:
            return (x*modu_expo((x*x)%m,(n-1)//2,m))%m
    # Function for modular multiplicative inverse via using extended_GCD_Euclid's
    def eEuclid(c, m):
        if m == 0:
            global d
            global x
            global y
            d = c
            x = 1
            y = 0
        else:
            eEuclid(m, c % m)
            temp = x
            x = y
            y = temp - (c // m) * y
    def mod_Inv(c, m):
        eEuclid(c, m)
        return (x % m + m) % m
    # Function for nCr%p using fermat (p is prime)
    def ncr(n, r, p):
        num = den = 1
        for i in range(r):
            num = (num * (n - i)) % p
            den = (den * (i + 1)) % p
        return (num * pow(den,
                          p - 2, p)) % p
    #*************************************************************
    for _ in range(int(input())):
        n,x=map(int,input().split())
        l=[int(y) for y in input().split()]
        e=0
        o=0
        for i in l:
            if i%2==0:
                e+=1
            else:
                o+=1
        ans='No'
        if o==0:
            print(ans)
            continue
        x1=1
        x2=x-1
        if x%2!=0:
            if x1 <= o and x2 <= e:
                ans = "Yes"
            while x1<=x and x>=0:
                if x1<=o and x2<=e:
                    ans="Yes"
                    break
                x1+=2
                x2-=2
        else:
            if x1 <= o and x2 <= e:
                ans = "Yes"
            while x1<=x-1 and x2>=1:
                if x1<=o and x2<=e:
                    ans="Yes"
                    break
                x1+=2
                x2-=2
        print(ans)

if __name__ == "__main__":
    main()