Problem: Omkar has received a message from Anton saying "Your story for problem A is confusing. Just make a formal statement." Because of this, Omkar gives you an array a = [a_1, a_2, …, a_n] of n distinct integers. An array b = [b_1, b_2, …, b_k] is called nice if for any two distinct elements b_i, b_j of b, |b_i-b_j| appears in b at least once. In addition, all elements in b must be distinct. Can you add several (maybe, 0) integers to a to create a nice array b of size at most 300? If a is already nice, you don't have to add any elements.

For example, array [3, 6, 9] is nice, as |6-3|=|9-6| = 3, which appears in the array, and |9-3| = 6, which appears in the array, while array [4, 2, 0, 6, 9] is not nice, as |9-4| = 5 is not present in the array.

For integers x and y, |x-y| = x-y if x > y and |x-y| = y-x otherwise.

Input

Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 50), the number of test cases. Description of the test cases follows.

The first line of each test case contains a single integer n (2 ≤ n ≤ 100) — the length of the array a.

The second line of each test case contains n distinct integers a_1, a_2, ⋅⋅⋅, a_n (-100 ≤ a_i ≤ 100) — the elements of the array a.

Output

For each test case, output one line containing YES if Omkar can create a nice array b by adding elements to a and NO otherwise. The case of each letter does not matter, so yEs and nO will also be accepted.

If the first line is YES, output a second line containing a single integer k (n ≤ k ≤ 300). 

Then output one line containing k distinct integers b_1, b_2, ⋅⋅⋅, b_k (-10^9 ≤ b_i ≤ 10^9), the elements of the nice array b. b_1, b_2, ⋅⋅⋅, b_k can be in any order. For each a_i in a, a_i must appear at least once in b.

It can be proved that if Omkar can create such an array b, then he can also do so in a way that satisfies the above constraints.

If multiple solutions exist, you can print any. 

Example

Input


4
3
3 0 9
2
3 4
5
-7 3 13 -2 8
4
4 8 12 6


Output


yes
4
6 0 3 9
yEs
5
5 3 1 2 4
NO
Yes
6
8 12 6 2 4 10

Note

For the first case, you can add integers to a to receive the array b = [6, 0, 3, 9]. Note that |6-3| = |9-6| = |3-0| = 3 and 3 is in b, |6-0| = |9-3| = 6 and 6 is in b, and |9-0| = 9 is in b, so b is nice.

For the second case, you can add integers to a to receive the array b = [5, 3, 1, 2, 4]. We have that |2-1| = |3-2| = |4-3| = |5-4| = 1 is in b, |3-1| = |4-2| = |5-3| = 2 is in b, |4-1| = |5-2| = 3 is in b, and |5-1| = 4 is in b, so b is nice.

For the fourth case, you can add integers to a to receive the array b = [8, 12, 6, 2, 4, 10]. We have that |4-2| = |6-4| = |8-6| = |10-8| = |12-10| = 2 is in b, |6-2| = |8-4| = |10-6| = |12-8| = 4 is in b, |8-2| = |10-4| = |12-6| = 6 is in b, |10-2| = |12-4| = 8 is in b, and |12-2| = 10 is in b, so b is nice.

It can be proven that for all other test cases it is impossible to create a nice array b.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Rearrange {

    public static void main(String args[]) throws IOException {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            int t = Integer.parseInt(br.readLine());
            while (t-- > 0) {
                int n = Integer.parseInt(br.readLine());
                String inp[] = br.readLine().split(" ");
                Set<Integer> set = new HashSet<Integer>(n);
                ArrayList<Integer> ll = new ArrayList<Integer>();
                boolean negative = false;
                for (int i = 0; i < inp.length; i++) {
                    set.add(Integer.parseInt(inp[i]));
                    ll.add(Integer.parseInt(inp[i]));
                    if (ll.get(i) < 0) {
                        negative = true;
                        break;
                    }
                }

                if (negative)
                    System.out.println("NO");
                else {
                    int k = n;
                    for (int i = 0; i < k && k <= 300; i++) {
                        int j = i + 1;
                        while (j < k) {
                            int value = Math.abs(ll.get(i) - ll.get(j));
                            if (!set.contains(value)) {
                                k++;
                                ll.add(value);
                                set.add(value);
                                i = -1;
                                break;
                            }
                            j++;
                        }
                    }
                    if (k > 300)    System.out.println("NO");
                    else {
                        System.out.println("YES\n" + k);
                        for (int val : ll)  System.out.print(val + " ");
                        System.out.println();
                    }
                }
            }
        }
    }
}