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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class qwe {
    static int p = 1000000007;
    static int power(int x, int y)//(x^y)%p
    {


        int res = 1;

        x = x % p;

        while (y > 0) {
            if (y % 2 == 1)
                res = (res * x) % p;
            y = y >> 1;
            x = (x * x) % p;
        }

        return res;
    }

    static int modInverse(int n)//mod inverse
    {
        return power(n, p - 2);
    }
    static int nCrModPFermat(int n, int r)//nCr
    {
        if (n<r)
            return 0;
        if (r == 0)
            return 1;
        int[] fac = new int[n + 1];
        fac[0] = 1;

        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i % p;

        return (fac[n] * modInverse(fac[r]) % p * modInverse(fac[n - r]) % p) % p;
    }

    static int gcd(int a, int b)
    {
        return (a % b == 0) ?
                Math.abs(b) : gcd(b,a%b);
    }
    static class pair
    {
        String first, second;
        public pair(String first, String second)
        {
            this.first = first;
            this.second = second;
        }
    }

    static String MEX(String s){
        String mex = "";
        boolean bo = true;
        HashSet<Character> set = new HashSet<>();
        for(int i=0;i<s.length();i++){
            set.add(s.charAt(i));
            if(set.size() == 26){

                mex += 'a';
                set.clear();


            }

        }

        for(char i= 'a';i<='z';i++){

            if(!set.contains(i)){
                mex+= i;
                return mex;
            }
        }
        return mex;
    }
    static ArrayList<Integer> count(String s, int n){
        ArrayList<Integer> arr = new ArrayList<>();
        int d = 0, k = 0;
        for(int i=0;i<n;i++){
            if(s.charAt(i) == 'D'){
                d++;

            }
            else{
                k++;
            }
            if(k==0){
                arr.add(d)
;
            }
            else if(d == 0){
                arr.add(k);
            }

            else if(d%k == 0 || k%d == 0){
                arr.add(Math.min(d,k));
            }
            else {
                arr.add(1);
            }
        }
        return arr;
    }
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

            int t = Integer.parseInt(br.readLine());
            while (t -->0){
                int N = Integer.parseInt(br.readLine());
                String s = br.readLine();
                String[] str = s.split(" ");
                ArrayList<Long> arr = new ArrayList<>();
                for(int i=0;i<N;i++)
                    arr.add(Long.parseLong(str[i]));
                Long max = Collections.max(arr);
                boolean bol = true;
                for(long i: arr){
                    if(i<0)
                    {
                        bol = false;
                        break;
                    }
                }
                String res = "";
                if(bol){
                    System.out.println("Yes");
                    System.out.println(max+1);
                    for(int i=0;i<=max;i++){
                        res += i+" ";
                    }
                    System.out.println(res);


                }
                else{
                    System.out.println("No");
                }



         }

        }


        }