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.Scanner;

public class D {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int inputs = in.nextInt();
		while(inputs-->0) {
			int n = in.nextInt();
			int[] nums = new int[n];
			for(int i = 0; i < n; i++) {
				nums[i] = Math.abs(in.nextInt());
			}
			boolean good = false;
			for(int i = 0; i < n; i++) {
				good |= check(0, i, 0, nums);
			}
			System.out.println(good ? "YES" : "NO");
		}
		in.close();
	}
	
	public static boolean check(int curr, int skip, int total, int[] nums) {
		if(total == nums[skip]) {
			return true;
		}
		if(curr >= nums.length) {
			return false;
		}
		if(curr == skip)
			return check(curr+1, skip, total, nums);
		return check(curr+1, skip, total, nums) || check(curr+1, skip, total+nums[curr], nums) || check(curr+1, skip, total-nums[curr], nums);
	}

}
