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:
"""
    This template is made by Satwik_Tiwari.
    python programmers can use this template :)) .
"""

#===============================================================================================
#importing some useful libraries.
from fractions import Fraction
import sys
import bisect
import heapq
from math import *
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned

#===============================================================================================
#some shortcuts

mod = 1000000007
def inp(): return sys.stdin.readline().strip() #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for p in range(t):
        solve()
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(a,b):
    ans = 1
    while(b>0):
        if(b%2==1):
            ans*=a
        a*=a
        b//=2
    return ans

#===============================================================================================
# code here ;))

def solve():
    n,x = sep()
    a = lis()
    pre = []
    for i in range(0,n):
        if(i==0):
            pre.append(a[i])
        else:
            pre.append(a[i]+pre[i-1])
    modarr = []
    for i in range(0,n):
        modarr.append(pre[i]%x)
    ans = 0
    help = -1
    for i in range(0,n):
        if(modarr[i] != 0):
            help = i
            break

    if(help == -1):
        print(-1)
        return

    for i in range(0,n):
        if(modarr[i] != 0):
            ans = max(ans,i+1)
        else:
            if(i > help):
                ans = max(ans,i-help)
    print(ans)





# testcase(1)
testcase(int(inp()))