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.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];
			for (int i = 0; i < n; i++)
				aa[i] = Math.abs(sc.nextInt());
			int[] log = new int[1 << n];
			for (int i = 0; i < n; i++)
				log[1 << i] = i;
			int[] ss = new int[1 << n];
			for (int p = 1; p < 1 << n; p++)
				ss[p] = ss[p & p - 1] + aa[log[p & -p]];
			int p_ = -1, q_ = -1;
			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]) {
						p_ = p; q_ = q;
						break;
					}
			if (p_ == -1) {
				println("NO");
				continue;
			}
			println("YES");
			print("0 " + ss[p_]);
			for (int p = p_ & p_ - 1; p > 0; p &= p - 1)
				print(" " + ss[p]);
			for (int q = q_ & q_ - 1; q > 0; q &= q - 1)
				print(" " + ss[q]);
			int r = (p_ | q_) ^ (1 << n) - 1;
			for (int i = 0; i < n; i++)
				if ((r & 1 << i) != 0)
					print(" " + aa[i]);
			println();
		}
	}
}
