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.
## necessary imports
import sys
input = sys.stdin.readline
from math import ceil, floor, factorial;

# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp

## gcd function
def gcd(a,b):
    if a == 0:
        return b
    return gcd(b%a, a)

## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
    if(k > n - k): 
        k = n - k 
    res = 1
    for i in range(k): 
        res = res * (n - i) 
        res = res / (i + 1) 
    return int(res) 

## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
    hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] < x:
            lo = mid+1
        else:
            hi = mid
    return lo

## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0 and n > 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0 and n > 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes

## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p): 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 

## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b

# find function with path compression included (recursive)
# def find(x, link):
#     if link[x] == x:
#         return x
#     link[x] = find(link[x], link);
#     return link[x];

# find function with path compression (ITERATIVE)
def find(x, link):
    p = x;
    while( p != link[p]):
        p = link[p];
    
    while( x != p):
        nex = link[x];
        link[x] = p;
        x = nex;
    return p;


# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x

## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime

#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e7 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array

################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve() 
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

## taking integer array input
def int_array():
    return list(map(int, input().strip().split()))
## taking string array input
def str_array():
    return input().strip().split();

#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');

################### ---------------- TEMPLATE ENDS HERE ---------------- ###################

for _ in range(int(input())):
    n, x = int_array(); a = int_array();
    if sum(a) % x != 0:
        print(n); continue;
    len1 = None; len2 = None; s = 0;
    for i in range(n):
        s += a[i];
        if s % x != 0:
            len1 = n - (i+1);
            break;
    s = 0; a = a[::-1];
    for i in range(n):
        s += a[i];
        if s % x != 0:
            len2 = n - (i+1);
            break;
    if len1 == None:
        print(-1);
    else:
        print(max(len1, len2));