Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.

An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

Input

The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.

The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.

The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.

Output

For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.

Example

Input


3
3 3
1 2 3
3 4
1 2 3
2 2
0 6


Output


2
3
-1

Note

In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.

In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.

In the third test case, all subarrays have an even sum, so the answer is -1.
def CeilIndex(A, l, r, key): 
  
    while (r - l > 1): 
      
        m = l + (r - l)//2
        if (A[m] >= key): 
            r = m 
        else: 
            l = m 
    return r 
   
def LongestIncreasingSubsequenceLength(A, size): 
  
    # Add boundary case, 
    # when array size is one 
   
    tailTable = [0 for i in range(size + 1)] 
    len = 0 # always points empty slot 
   
    tailTable[0] = A[0] 
    len = 1
    for i in range(1, size): 
      
        if (A[i] < tailTable[0]): 
  
            # new smallest value 
            tailTable[0] = A[i] 
   
        elif (A[i] > tailTable[len-1]): 
  
            # A[i] wants to extend 
            # largest subsequence 
            tailTable[len] = A[i] 
            len+= 1
   
        else: 
            # A[i] wants to be current 
            # end candidate of an existing 
            # subsequence. It will replace 
            # ceil value in tailTable 
            tailTable[CeilIndex(tailTable, -1, len-1, A[i])] = A[i] 
          
   
    return len
def countDistinct(arr, n): 
  
    # First sort the array so that all 
    # occurrences become consecutive 
    arr.sort(); 
  
    # Traverse the sorted array 
    res = 0; 
    i = 0; 
    while(i < n): 
  
        # Move the index ahead while 
        # there are duplicates 
        while (i < n - 1 and 
               arr[i] == arr[i + 1]): 
            i += 1; 
  
        res += 1; 
        i += 1; 
  
    return res; 
for _ in range(int(input())):
	n = int(input())
	arr = [int(s) for s in input().split()]
	x = LongestIncreasingSubsequenceLength(arr,len(arr))
	c  = countDistinct(arr,len(arr))
	print(max(x,c))

