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:
//WHEN IN DOUBT , USE BRUTE FORCE !!!!!!!!!
import java.io.*;
import java.util.*;
//class Main    //AtCoder
//class Solution // Codechef
public class Solution2 //Codeforces
{
    public static void main(String args[])
    {
        try {
            FastReader sc = new FastReader();
            int TT = sc.nextInt();
            for(int hehe=1 ; hehe  <= TT ; hehe++) {
                int N=sc.nextInt();
                int arr[]=new int[N];
                for(int i=0;i<N;i++)
                    arr[i]=sc.nextInt();
                ruffleSort(arr);
                if(arr[0]<0){
                    sopln("NO");
                }else{
                    sopln("YES");
                    sopln(1+arr[N-1]);
                    for(int i=0;i<=arr[N-1];i++){
                        sop(i+" ");
                    }
                    sopln("");
                    //sopln("YES");
                }
            }
        }catch(Exception e){
            sopln(e.toString());
        }
    }

    private static int isCheck(long[] arr) {
        boolean b=true;
        long diff=arr[1]-arr[0];
        for(int i=1;i<arr.length;i++){
            if((arr[i]-arr[i-1])!=diff)
                return 0;
        }
        return (diff>0)?1:-1;
    }

    static void ruffleSort(int[] a) {
        ArrayList<Integer> l=new ArrayList<>();
        for (int i:a) l.add(i);
        Collections.sort(l);
        for (int i=0; i<a.length; i++) a[i]=l.get(i);
    }

    static void ruffleSortRev(long[] a) {
        ArrayList<Long> l=new ArrayList<>();
        for (long i:a) l.add(i);
        Collections.sort(l);Collections.reverse(l);
        for (int i=0; i<a.length; i++) a[i]=l.get(i);
    }

    static int power(int x, int y, int p)
    {

        // Initialize result
        int res = 1;

        // Update x if it is more than or
        // equal to p
        x = x % p;

        while (y > 0) {

            // If y is odd, multiply x
            // with result
            if (y % 2 == 1)
                res = (res * x) % p;

            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }

        return res;
    }

    // Returns n^(-1) mod p
    static int modInverse(int n, int p)
    {
        return power(n, p - 2, p);
    }

    // Returns nCr % p using Fermat's
    // little theorem.
    static int nCrModPFermat(int n, int r,
                             int p)
    {

        if (n<r)
            return 0;
        // Base case
        if (r == 0)
            return 1;

        // Fill factorial array so that we
        // can find all factorial of r, n
        // and n-r
        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)
                % p * modInverse(fac[n - r], p)
                % p)
                % p;
    }


    static boolean isPrime(long n)
    {

        // Check if number is less than
        // equal to 1
        if (n <= 1)
            return false;

            // Check if number is 2
        else if (n == 2)
            return true;

            // Check if n is a multiple of 2
        else if (n % 2 == 0)
            return false;

        // If not, then just check the odds
        for (int i = 3; i <= Math.sqrt(n); i += 2)
        {
            if (n % i == 0)
                return false;
        }
        return true;
    }

    static long gcd(long a, long b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static long countDig(long N){
        if(N < 10)
            return 1;
        else
            return 1+countDig(N/10);
    }


    public final static int d = 256;
    static int MOD = 1000000007;
    static final double PI = Math.PI;
    private static BufferedReader in = new BufferedReader (new InputStreamReader (System.in));
    private static PrintWriter out = new PrintWriter (new OutputStreamWriter (System.out));
    static class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader()
        {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                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());
        }

        char nextChar()
        {
            try {
                return (char) (br.read());
            }catch (IOException e){
                return '~';
            }
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
    static void sop(Object o){System.out.print(o);}
    static double ceil(double d){
        return Math.ceil(d);
    }
    static double floor(double d){
        return Math.floor(d);
    }
    static double round(double d){
        return Math.round(d);
    }

    static void sopln(Object o){System.out.println(o);}
    static void printArray(boolean[] arr){
        for(int i=0;i<arr.length;i++){
            sop(arr[i]+" ");
        }
        sopln("");
    }
    static void printArray(long[] arr){
        for(int i=0;i<arr.length;i++){
            sop(arr[i]+" ");
        }
        sopln("");
    }
    static long power(long x, long y, long p)
    {

        // Initialize result
        long res = 1;

        // Update x if it is more than or
        // equal to p
        x = x % p;

        while (y > 0) {

            // If y is odd, multiply x
            // with result
            if (y % 2 == 1)
                res = (res * x) % p;

            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }

        return res;
    }

    // Returns n^(-1) mod p
    static long modInverse(long n, int p)
    {
        return power(n, p - 2, p);
    }

    // Returns nCr % p using Fermat's
    // little theorem.

    ///

    // Function to find modular
// inverse of a under modulo p
// using Fermat's method.
// Assumption: p is prime
    static long modFact(int n,
                        int p)
    {
        if (n >= p)
            return 0;

        long result = 1;
        for (int i = 1; i <= n; i++)
            result = (result * i) % p;

        return result;
    }

}