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.io.*;
  import java.util.*;
  
  public class CF1552D extends PrintWriter {
  	CF1552D() { super(System.out); }
  	Scanner sc = new Scanner(System.in);
  	public static void main(String[] $) {
  		CF1552D o = new CF1552D(); o.main(); o.flush();
  	}
  
  	void main() {
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			int[] aa = new int[n];
  			boolean zero = false;
  			for (int i = 0; i < n; i++) {
  				aa[i] = Math.abs(sc.nextInt());
  				if (aa[i] == 0)
  					zero = true;
  			}
  			if (zero) {
  				println("YES");
  				continue;
  			}
  			if (n == 1) {
  				println("NO");
  				continue;
  			}
  			int[] ss = new int[1 << n];
  			for (int i = 0; i < n; i++)
  				ss[1 << i] = aa[i];
  			for (int p = 1; p < 1 << n; p++)
  				ss[p] = ss[p & p - 1] + ss[p & -p];
  			boolean yes = false;
  			for (int p = 1; p < 1 << n; p++)
  				for (int q = p + 1; q < 1 << n; q++)
  					if ((p & q) == 0 && ss[p] == ss[q]) {
  						yes = true;
  						break;
  					}
  			println(yes ? "YES" : "NO");
  		}
  	}
  }
  