description: You are given a sequence of n integers a_1,   a_2,   ...,   a_n.
  
  Does there exist a sequence of n integers b_1,   b_2,   ...,   b_n such that the following property holds?
  
    * For each 1 ≤ i ≤ n, there exist two (not necessarily distinct) indices j and k (1 ≤ j,   k ≤ n) such that a_i = b_j - b_k. 
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 20) — the number of test cases. Then t test cases follow.
  
  The first line of each test case contains one integer n (1 ≤ n ≤ 10).
  
  The second line of each test case contains the n integers a_1,   ...,   a_n (-10^5 ≤ a_i ≤ 10^5).
  
  Output
  
  For each test case, output a line containing YES if a sequence b_1,   ...,   b_n satisfying the required property exists, and NO otherwise.
  
  Example
  
  Input
  
  
  5
  5
  4 -7 -1 5 10
  1
  0
  3
  1 10 100
  4
  -3 2 10 2
  9
  25 -171 250 174 152 242 100 -205 -258
  
  
  Output
  
  
  YES
  YES
  NO
  YES
  YES
  
  Note
  
  In the first test case, the sequence b = [-9,   2,   1,   3,   -2] satisfies the property. Indeed, the following holds: 
  
    * a_1 = 4 = 2 - (-2) = b_2 - b_5; 
    * a_2 = -7 = -9 - (-2) = b_1 - b_5; 
    * a_3 = -1 = 1 - 2 = b_3 - b_2; 
    * a_4 = 5 = 3 - (-2) = b_4 - b_5; 
    * a_5 = 10 = 1 - (-9) = b_3 - b_1. 
  
  
  
  In the second test case, it is sufficient to choose b = [0], since a_1 = 0 = 0 - 0 = b_1 - b_1.
  
  In the third test case, it is possible to show that no sequence b of length 3 satisfies the property.
solution: import java.util.*;
  import java.io.*;
  public class D {
  	static ArrayList<Integer> set = new ArrayList<Integer>();
  	static int[] a;
  	static int test;
  	public static void main(String[] args) throws IOException{
  		Scanner in = new Scanner(System.in);
  		int t = in.nextInt();
  		for(int q=0;q<t;q++) {
  			int n = in.nextInt();
  			a = new int[n];
  			for(int i=0;i<n;i++) {
  				a[i] = Math.abs(in.nextInt());
  			}
  			boolean yes = false;
  			for(int i=0;i<n;i++) {
  				if(a[i]==0) yes = true;
  				int[] b = new int[n-1];
  				int index=0;
  				for(int j=0;j<n;j++) {
  					if(j!=i) {
  						b[index] = a[j];
  						index++;
  					}
  				}
  				test = a[i];
  				set = new ArrayList<Integer>();
  				if(b.length!=0) {
  					if(subset(b,0)) yes = true;					
  				}
  			}
  			if(yes) {
  				System.out.println("YES");
  			}else {
  				System.out.println("NO");
  			}
  		}
  	}
  	
  	static boolean subset(int[] a,int i) {
  //		System.out.println(set);
  		if(i>=a.length) {
  			if(binary(new int[set.size()],0)) return true;
  			return false;
  		}
  		set.add(a[i]);
  		if(subset(a,i+1))return true;
  		set.remove(set.size()-1);
  		if(subset(a,i+1))return true;
  		return false;
  	}
  	
  	static boolean binary(int[] b,int i) {
  		if(i==b.length) {
  			int sum=0;
  			for(int j=0;j<b.length;j++) {
  				if(b[j]==0) {
  					sum+= set.get(j);
  				}else {
  					sum-= set.get(j);
  				}
  			}
  //			System.out.println(sum+" "+test);
  			if(sum==test) {
  				return true;
  			}else {
  				return false;
  			}
  		}
  		b[i] = 0;
  		if(binary(b,i+1))return true;
  		b[i] = 1;
  		if(binary(b,i+1))return true;
  		return false;
  	}
  	
  }
  
  //Is only possible if you can use at least 1 number in b for at least 2 numbers in a
  
  