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.
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

