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.*;
public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int n  =sc.nextInt();
            int[] arr = new int[n];
            for(int i=0;i<n;i++){
                arr[i] =sc.nextInt();
            }
            int sum=0;
            String ans="";
            boolean zero = true;
            for(int i=0;i<n;i++){
                if(arr[i]!=0){
                    zero = false;
                    break;
                }
            }
            if(zero) ans = "Yes";
            else{
                boolean flag1 = true;
                boolean flag2 = true;
                 //inc
                int max = Math.max(Integer.MIN_VALUE, arr[0]);
                int mx = max;
                for(int i=1;i<n;i++){
                    mx = Math.max(arr[i], mx);
                    if(mx!=arr[i]){
                        flag1 = false;
                        break;
                    }
                }
                //dec
                int min = Math.min(Integer.MAX_VALUE, arr[0]);
                int mn = min;
                for(int i=1;i<n;i++){ 
                    mn = Math.min(arr[i], mn);
                    if(mn!=arr[i]){
                        flag2 = false;
                        break;
                    }
                }

                if(flag1 || flag2){
                    ans="NO";
                }
                else
                    ans="Yes";
            }
            System.out.println(ans);
        }
    }
}