Ehab has an array a of length n. He has just enough free time to make a new array consisting of n copies of the old array, written back-to-back. What will be the length of the new array's longest increasing subsequence?

A sequence a is a subsequence of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements. The longest increasing subsequence of an array is the longest subsequence such that its elements are ordered in strictly increasing order.

Input

The first line contains an integer t — the number of test cases you need to solve. The description of the test cases follows.

The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) — the number of elements in the array a.

The second line contains n space-separated integers a_1, a_2, …, a_{n} (1 ≤ a_i ≤ 10^9) — the elements of the array a.

The sum of n across the test cases doesn't exceed 10^5.

Output

For each testcase, output the length of the longest increasing subsequence of a if you concatenate it to itself n times.

Example

Input


2
3
3 2 1
6
3 1 4 1 5 9


Output


3
5

Note

In the first sample, the new array is [3,2,1,3,2,1,3,2,1]. The longest increasing subsequence is marked in bold.

In the second sample, the longest increasing subsequence will be [1,3,4,5,9].

import java.util.*;

import java.io.*;
import  java.lang.Math.*;
public class Solutions {
	 
static int MAX=Integer.MAX_VALUE;
		static int MIN=Integer.MIN_VALUE;
//static ArrayList<ArrayList<Integer>>list=new ArrayList<ArrayList<Integer>>();
	static FastScanner scr=new FastScanner();
	static PrintStream out=new PrintStream(System.out);
//	static ArrayList<Integer>list[];
//	static StringBuilder sb=new StringBuilder();
	public static void main(String []args) {
		int t=scr.nextInt();
		while(t-->0) {
			solve();
		}
//		out.println(sb);
	}
//	static HashMap<Integer,Integer>hm;
	static void solve() {
		int n=scr.nextInt();
		HashSet<Integer>hs=new HashSet<>();
		for(int i=0;i<n;i++) {
			hs.add(scr.nextInt());
		}
		out.println(hs.size());
	}
	static class pair{
		int x;
		int y;
		pair(int x,int y){
			this.x=x;
			this.y=y;
		}
	}
	static boolean isVowel(char c) {
		return c=='a' || c=='e'||c=='i'|| c=='o' || c=='u';
	}
	static int gcd(int a,int b){
		if(b==0) {
			return a;
		}
		
		return gcd(b,a%b);
	}
	static long modPow(long base,long exp,long mod) {
		if(exp==0) {
			return 1;
		}
		if(exp%2==0) {
			long res=(modPow(base,exp/2,mod));
			return (res*res)%mod;
		}
		return (base*modPow(base,exp-1,mod))%mod;
	}
	static long gcd(long a,long b) {
		if(b==0) {
			return a;
		}
		return gcd(b,a%b);
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
	 	String next() {
	 		while (!st.hasMoreTokens())
	 			try {
	 				st=new StringTokenizer(br.readLine());
	 			} catch (IOException e) {
	 				e.printStackTrace();
	 			}
	 			return st.nextToken();
		}
		int[] sort(int a[]) {
			Arrays.sort(a);
			return a;
		}
		int []reverse(int a[]){
			int b[]=new int[a.length];
			int index=0;
			for(int i=a.length-1;i>=0;i--) {
				b[index]=a[i];
			}
			return b;
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long[] readLongArray(int n) {
			long  [] a=new long  [n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}

}
