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 atexit, io, sys 
def retmax(a):
    s=0;n=a[0]
    for j in range(1,len(a)):
        if n<a[j]:
            s=j;n=a[j]
    return s
from math import log,floor
# A stream implementation using an in-memory bytes  
# buffer. It inherits BufferedIOBase. 
buffer = io.BytesIO() 
sys.stdout = buffer
  
# print via here 
@atexit.register 
def write(): 
    sys.__stdout__.write(buffer.getvalue())
for _ in range(input()):
    n,k=map(int,raw_input().split())
    kv=[0 for j in range(100)]
    a=map(int,raw_input().split())
    ke=1;p=1;
    s=[]
    i=1
    while i<=(10**16):
        s.append(i);i*=k
    s.append(i)
    while True:
        q=0
        for j in range(n):
            if a[j]>0:
                q=1;break
        if q==0:
            break
        v=retmax(a)
        
        for i in range(len(s)):
            if a[v]<s[i]:
                g=i-1;break
       
        if kv[g]:
            p=0;break
        else:
            kv[g]=1;a[v]=int(a[v]-pow(k,g))
            
    if p:
        print"YES"
    else:
        print"NO"