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);
        StringBuilder out = new StringBuilder();
        int t = sc.nextInt();
        while (t-- > 0) {
            int[] a = new int[sc.nextInt()];
            for (int i = 0; i < a.length; i++) {
                a[i] = sc.nextInt();
            }
            solve(a, out);
        }
        System.out.print(out.toString());
    }

    static void solve(int[] a, StringBuilder out) {
        if (a.length == 1) {
            if (a[0] == 0) {
                out.append("YES\n");
            } else {
                out.append("NO\n");
            }
            return;
        }
        Set<Integer> set = new HashSet<>();
        set.add(0);
        for (int i = 0; i < a.length - 1; i++) {
            int[] va = new int[set.size()];
            int vai = 0;
            for (Integer v : set) {
                va[vai++] = v;
            }
            for (int j = 0; j < va.length; j++) {
                set.add(va[j] + a[i]);
                set.add(va[j] - a[i]);
            }
        }
        if (set.contains(a[a.length - 1])) {
            out.append("YES\n");
        } else {
            out.append("NO\n");
        }
    }
}