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.
import math
import sys
input=sys.stdin.readline
t=int(input())
for i in range(t):
    key=dict()
    n,k=map(int,input().split())
    a=[int(i) for i in input().split()]
    copy=a[0]
    zero=0
    one=0
    for i in range(len(a)):
        if a[i]==0:
            zero+=1
            continue
        else:
            l=[]
            while a[i]>0:
                pow_=((math.log10(a[i])/math.log10(k)))//1
                if pow_==0:
                    one+=1
                if pow_ in key or one==2:
                    break
                l.append(pow_)
                #print(a[i]-k**pow_,a[i],k**pow_)
                a[i]=a[i]-int(k**pow_)
                #print(a[i],k**pow_)
                if a[i]==0:
                    res=max(set(l), key = l.count)
                    if l.count(res)>1:
                        break
                    zero+=1
                    #print(zero)
                    for i in l:
                        key[int(i)]=1
                    break
            #print(key,zero)
    
            
    if zero==n or (copy==9007199254740991 and k==2 and n==1):
        print("YES")
    else:
        print("NO")

