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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.*;
import java.io.*; 
import java.util.HashSet;

import javax.lang.model.util.ElementScanner6;

  
 public class STRING{
     
      public static void main( String args[]) {  
         
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); 
       Scanner sc = new Scanner(System.in);  
      PrintWriter writer = new PrintWriter(System.out);
      int t = sc.nextInt();
      while(t-->0){
          int n = sc.nextInt();
          int x = sc.nextInt();
          int arr[] = new int [n];
          int arr1[] = new int[n];
          int sum=0;
          for(int i=0;i<n;i++){
              arr[i] = sc.nextInt();
              sum+=arr[i];
              if(arr[i]%x==0)
               arr1[i]=1;
    
          }
          int max =-1,min=1000000;
          for(int i=0;i<n;i++){
              if(arr1[i]==0){
                min = Math.min(i,min);
                max = Math.max(i,max);
              }
          } 
          if(sum%x!=0)
           System.out.println(n);
          else if(max==-1&&min==1000000)
           System.out.println(-1);
          else 
           System.out.println(Math.max(n-min-1,max));   
            
        
          
        
      }
      
      
        
}  

   public static String replace(String str, int index, char replace){     
    if(str==null){
        return str;
    }else if(index<0 || index>=str.length()){
        return str;
    }
    char[] chars = str.toCharArray();
    chars[index] = replace;
    return String.valueOf(chars);       
}
public static HashSet<Integer> primeFactors(int n) 
{   
    HashSet<Integer> h = new HashSet<Integer>();
    // Print the number of 2s that divide n 
    while (n%2==0) 
    { 
        h.add(2);
        n /= 2; 
    } 


    // n must be odd at this point.  So we can 
    // skip one element (Note i = i +2) 
    for (int i = 3; i <= Math.sqrt(n); i+= 2) 
    { 
      
        // While i divides n, print i and divide n 
        while (n%i == 0) 
        {  
            h.add(i);  
            n /= i; 
        } 
    } 
     
    // This condition is to handle the case whien 
    // n is a prime number greater than 2 
    if(n>2)
      h.add(n);

   return h ;    
     
 }
public static boolean isPrime(int n) 
    { 
        // Corner cases 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
  
        // This is checked so that we can skip 
        // middle five numbers in below loop 
        if (n % 2 == 0 || n % 3 == 0) 
            return false; 
  
        for (int i = 5; i * i <= n; i = i + 6) 
            if (n % i == 0 || n % (i + 2) == 0) 
                return false; 
  
        return true; 
    } 

}