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().strip())
kmax = 10**16 + 10

def find_possible(num,k,power_array):
    start = 1
    power = 0
    if num == 1:
        power_array[0] += 1
        return 
    while True:
        
        result = start * k
        if result > num:
            break
        else:
            start = result
            power += 1
            if start == num:
                power_array[power] += 1
                return 
    high = power
    
    
    while True:
        result = num
        rep = 0
        for i in range(0,k):
            if result - start == 0:
                rep += 1
                result = 0
                break
            elif result - start < 0:
                break
            result -= start
            rep += 1
        #print('num:{},result:{},start:{}'.format(num,result,start))
        if result == 0:
            for i in range(rep):
                power_array[power] += 1
            return high
        if rep > 0:
            for i in range(rep):
                power_array[power] += 1
            power -= 1
            start = start//k
            num = result
            
        else:
            power -= 1
            start = start//k
            
            
for _ in range(t):
    power_array = [0 for i in range(100)]
    n,k = map(int,input().strip().split())
    #n = int(input().strip())
    nums = [int(i) for i in input().strip().split()]
    nums.sort()
    has = True
    cur_index = 0
    while cur_index < n:
        if nums[cur_index] == 0:
            cur_index += 1
        else:
            result = find_possible(nums[cur_index],k,power_array)
            cur_index += 1

    for i in power_array:
        if i >= 2:
            has = False
            break
    if has:
        print('YES')
    else:
        print('NO')
