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.
#    Unmesh Kumar Gupta
#    CSE'22,IIT Roorkee

from math import *

mod=1000000007
inf=10000000000000000

def binpow(a,b):
    global mod
    if b==1:
        return a%mod
    u=binpow(a,b//2)
    if b%2==0:
        return (u*u)%mod
    else:
        return (u*u*a)%mod

def gcd(a,b):
    if b==0:
        return a
    return gcd(b,a%b)



temp=[0]*500
used=[0]*500


def bs(low,high,target):
    if low==high:
        return low
    elif low+1==high:
        if temp[high]>target:
            return low
        else:
            return high
    else:
        mid=(low+high)//2
        if temp[mid]==target:
            return mid
        elif temp[mid]>target:
            return bs(low,mid-1,target)
        else:
            return bs(mid,high,target)



for _ in range(int(input())):

    n,k=[int(i) for i in input().split()]
    for i in range(300):
        used[i]=0
    temp[0]=1
    counter=1
    while(temp[counter-1]<pow(10,20)):
        temp[counter]=temp[counter-1]*k
        counter+=1
    arr=[int(i) for i in input().split()]
    b=True
    # print(counter)
    # for i in range(counter):
    #     print(temp[i])
    for a in arr:
        # print(" a ",a)
        while a>0:
            # print(" a ",a)
            g=bs(0,counter-2,a)
            # print("g  ",g)
            if used[g]==1:
                b=False
                break
            else:
                a-=temp[g]
                used[g]=1
        if not b:
            break
    if not b:
        print("NO")
    else:
        print("YES")
