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.
## python 3
from math import log

MAX_EXP = 0
used = []

def getExpo(x, k):
    global MAX_EXP
    l, r = 0, MAX_EXP
    ##
    while l <= r:
        mid = (l + r + 1)//2
        if x % k**mid == 0 and x % k**(mid+1) != 0:
            return mid
        if x % k**mid == 0:
            l = mid
        else:
            r = mid - 1
    ##
    return -1

def test(x, k):
    global used
    D = {}

    cumExp = 0

    ##
    while x > 0:
        exp = getExpo(x, k)
        if exp == -1:
            return False
        # print("x = %d | exp = %d\n" % (x, exp))
        ##
        x //= (k ** exp)
        x -= 1
        ##
        cumExp += exp
        D[cumExp] = D.get(cumExp, 0) + 1
        if D[cumExp] > 1:
            return False

    ## Check for collision
    for k in D.keys():
        if used[k] == 1:
            return False
    ## Comfirmation
    for k in D.keys():
        used[k] = 1
    return True
        

def solve():
    n, k = map(int, input().split())
    a    = list(map(int, input().split()))
    ###
    global used
    used = [0] * 64
    global MAX_EXP
    MAX_EXP = int(log(max(a) + 1, k)) + 5
    # print("MAX_EXP = ", MAX_EXP)
    ###
    for x in a:
        # print("$Testing ", x);
        if test(x, k) == False:
            print("NO\n")
            return
    ###
    print("YES\n")

t = int(input())
while t > 0:
    t -= 1
    ##
    solve()
