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.util.*;
import java.lang.*;
import java.io.*;
public class Dominated_Subarray {

  public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

  static class FastReader {

    BufferedReader br;
    StringTokenizer st;

    public FastReader() {
      br = new BufferedReader(new
          InputStreamReader(System.in));
    }

    String next() {
      while (st == null || !st.hasMoreElements()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }

    int nextInt() {
      return Integer.parseInt(next());
    }

    long nextLong() {
      return Long.parseLong(next());
    }

    double nextDouble() {
      return Double.parseDouble(next());
    }

    String[] nextSArray() {
      String sr[] = null;
      try {
        sr = br.readLine().trim().split(" ");
      } catch (IOException e) {
        e.printStackTrace();
      }
      return sr;
    }

    String nextLine() {
      String str = "";
      try {
        str = br.readLine();
      } catch (IOException e) {
        e.printStackTrace();
      }

      return str;
    }
  }

  //  static int MOD=1000000000+7;
//  public static void sortbyCol(int arr[][],int col)
//  {
//    Arrays.sort(arr,new Comparator<int[]>(){
//      @Override
//      public int compare(int[] o1, int[] o2) {
//        // To sort in descending order revert
//        // the '>' Operator
//        if(o1[col]>o2[col])
//          return 1;
//        else
//          return -1;
//      }
//    });
//  }


  public static void main(String[] args) throws java.lang.Exception {
    FastReader sc = new FastReader();
    int t = sc.nextInt();
    for (int t1 = 0; t1 < t; ++t1) {
      int n = sc.nextInt();
      int x = sc.nextInt();
      boolean found = false;
      int indexa=0,indexb=0;
      long sum = 0;
      for (int i = 0; i < n; ++i) {
        int ele = sc.nextInt();
        if (ele % x != 0)
        {
          if(!found)
          {
            indexa=i;
            indexb=i;
          }
          found=true;
          indexb=i;
        }
        sum += ele;

      }
      if (sum % x != 0)
        out.println(n);
      else if (found)
      {
        int ans=Math.max(Math.max(indexa-0,n-indexa-1),Math.max(indexb-0,n-indexb-1));
       // out.println(indexa+" "+indexb);
        out.println(ans);
      }
      else
        out.println(-1);

    }
    out.close();
  }


  }