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.*;
import static java.lang.System.in;
import static java.lang.System.out;

public class A {    
    static boolean correct(ArrayList<Integer> arr, TreeSet<Integer> tree){
        for(int i = 0; i < arr.size(); i++){
            for(int j = i+1; j < arr.size(); j++){
                int one = Math.abs(arr.get(i) - arr.get(j));
                if(!tree.contains(one)) return false;
            }
        }
        return true;
    }
    public static void main(String args[]) throws Exception {   
        Foster sc = new Foster();
        PrintWriter p = new PrintWriter(out);
        int t = sc.nextInt();
        while(t--!=0){
            int n = sc.nextInt();
            int a[] = sc.intArray(n);
            a = sort(a);
            boolean neg = false, pos = false, zero = false;
            for(int i : a){
                if(i < 0)   neg = true;
                if(i > 0)   pos = true;
                if(i == 0)  zero = true;
            }
            if(n==1){
                p.println("YES");
                p.println("1");
                p.println(a[0]);
            }
            else if(neg){
                p.println("NO");
            }
            else {
                p.println("YES");
                p.println(300);
                int start = 1;
                if(zero)    start = 0;
                for(int i = start; i < start+300; i++){
                    p.print(i + " ");
                }
                p.println();
            }
        }
        p.close();
    }
    
    static long[] sort(long a[]){
        ArrayList<Long> arr = new ArrayList<>();
        for(long i : a){
            arr.add(i);
        }
        Collections.sort(arr);
        for(int i = 0; i < arr.size(); i++){
            a[i] = arr.get(i);
        }
        return a;
    }

    static int[] sort(int a[]){
        ArrayList<Integer> arr = new ArrayList<>();
        for(int i : a){
            arr.add(i);
        }
        Collections.sort(arr);
        // Collections.reverse(arr);
        for(int i = 0; i < arr.size(); i++){
            a[i] = arr.get(i);
        }
        return a;
    }
    
/*
 * Check overflow in pow function or in general
 * Check indices of read array functions
 * Think of an easier solution because the problems you solve are always easy
 * Check the iterator of loop
 */      

    static class Foster {
        BufferedReader br = new BufferedReader(new InputStreamReader(in));
        StringTokenizer st = new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } 
                catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        int[] intArray(int n) {
            int arr[] = new int[n];
            for(int i = 0; i < n; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }
        long[] longArray(int n) {
            long arr[] = new long[n];
            for(int i = 0; i < n; i++) {
                arr[i] = nextLong();
            }
            return arr;
        }
        static long pow(long... a) {
            long mod = Long.MAX_VALUE;
            if(a.length == 3)   mod = a[2];
            long res = 1;
            while(a[1] > 0) {
                if((a[1] & 1) == 1)
                    res = (res * a[0]) % mod;
                a[1] /= 2;
                a[0] = (a[0] * a[0]) % mod;
            }
            return res;
        }
    }
}