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 MainClass {
    public static void main(String[] args) {
        Reader in = new Reader(System.in);
        StringBuilder stringBuilder = new StringBuilder();
        int t = in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = in.nextLong();
            }
            boolean ff = false;
            for (int i = 1; i < (1 << n); i++) {
                for (int j = 0; j < (1 << solve(i)); j++) {
                    int sum = 0;
                    int count = 0;
                    for (int bit = 0; bit < n; bit++) {
                        int mask = 1 << bit;
                        int sign = 1;
                        if ((j & (1 << count)) > 0) {
                            sign = -1;
                        }
                        if ((i & mask) > 0) {
                            sum += sign * a[bit];
                            count++;
                        }
                    }
                    if (sum == 0) {
                        ff = true;
                    }
                }
            }
            if (ff) {
                stringBuilder.append("YES\n");
            } else {
                stringBuilder.append("NO\n");
            }
        }
        System.out.println(stringBuilder);
    }
    public static int solve(int x) {
        int count = 0;
        while (x > 0) {
            x = x & (x - 1);
            count++;
        }
        return count;
    }
}
class Reader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public Reader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

}
/*
If atleast one zero found, then b = a => YES
Else if a_i - a_j = a_k, then use b_k = 0 => YES
Else if duplicates in a, then YES

1 2 5

 */