Problem: 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.

Solution (in PYTHON3):

Solution:
#qn given an arr of size n for every  i position find a min j such that j>i and arr[j]>arr[i]
#it prints the position of the nearest .
import sys
input = sys.stdin.readline
 
############ ---- Input Functions ---- #######Start#####
def inp():
	return(int(input()))
def inlt(): 
	return(list(map(int,input().split())))
def insr():
	s = input()
	return(list(s[:len(s) - 1]))
def invr():
	return(map(int,input().split()))
 ############ ---- Input Functions ---- #######End#####   
def bin_search(arr,l,r,val):#strickly greater
	if arr[r] <= val:
		return r+1
	if r-l < 2:
		if arr[l]>val:
			return l
		else:
			return r
	mid = int((l+r)/2)
	if arr[mid] <= val:
		return bin_search(arr,mid,r,val)
	else:
		return bin_search(arr,l,mid,val)

def pr_list(a):
	print(*a, sep=" ")
def main():
	tests =  inp()
	for test in range(tests):
		n = inp()
		a = inlt()
		ans = 0
		hashm = {}
		for i in a:
			if i not in hashm:
				ans +=1
				hashm[i]=1
		print(ans)
		
	





	return 0


# if for two position i,j such that j<i and arr[j]>arr[i] I wont need to store the position i as I wont requre it for any further computation. I cound simply use i instead of j.
#to do that i main tain a stack which is initiallized with the last elment and I traverse the arr backwords because for any ith entry the answer does not depends on 0 to i-1.
# for every element I would continue to pop the stack (from last) till i recieve an elemnt larger than current one. The elemnt which i poped are not needed as i could use current elemnt.
# the order is n as an element only enters the stack ones.
if __name__== "__main__":
  main()