Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can: 

  * either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i; 
  * or not choose any position and skip this step. 



You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?

Input

The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.

The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.

Output

For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.

Example

Input


5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810


Output


YES
YES
NO
NO
YES

Note

In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.

In the second test case, you can add k^0 to v_1 and stop the algorithm.

In the third test case, you can't make two 1 in the array v.

In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2.
t=int(input(""))
while(t>0):
    n,k=input("").split()
    n=int(n)
    k=int(k)
    g=[]
    l=list(map(int,input().split()))
    d=max(l)
    j=0
    e=0
    c=0
    while(k**j<=d):
        e=j
        j+=1
    for i in range(0,len(l)):
        s=""
        for j in range(e+1,-1,-1):
            if(k**j<=l[i]):
                l[i]=l[i]-k**j
                s=s+"1"
            else:
                s=s+"0"
            if(l[i]==0):
                for q in range(j-1,-1,-1):
                    s=s+"0"
                g.append(s)
                break
        if(l[i]>0):
            print("NO")
            c=1
            break
    if(c==0):
        sum=0
        u=0
        for i in g:
            sum=sum+int(i)
            h=str(sum)
            for i in h:
                if(int(i)>1):
                    print("NO")
                    c=1
                    u=1
                    break
            if(u==1):
                break
        if(c==0):
            print("YES")
    t=t-1
                
