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 Main {
    /************************ SOLUTION STARTS HERE ************************/

    // Recursive function to return gcd of a and b
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    private static void solve() {
        int t = nextInt();
        while(t-- > 0) {
            int n = nextInt();
            List<Integer> list = new ArrayList<>();
            for(int j =0;j<n;j++) {
                Integer c = nextInt();
                list.add(c);
            }

            Set<Integer> set = new HashSet<>(list);
            int i = 0;
            boolean done = false;
            while(i <= 300) {
                int mx = 0;
                for(int j=0;j<list.size();j++)
                    mx = Math.max(mx, Math.abs(list.get(j)));

                List<Integer> newOnes = new ArrayList<>();
                boolean bigNo = false;
                boolean newAdded = false;
                for(int j=0;j<list.size();j++) {
                    for(int k=j+1;k<list.size();k++) {
                        int cur = Math.abs(list.get(j) - list.get(k));
                        if(set.contains(cur))
                            continue;
                        else {
                            if(cur > mx) {
                                bigNo = true;
                                break;
                            }
                            newOnes.add(cur);
                            set.add(cur);
                            newAdded = true;
                        }
                    }
                    if(bigNo)
                        break;
                }
                if(bigNo)
                    break;
                if(newAdded) {
                    for(int j=0;j<newOnes.size();j++) {
                        list.add(newOnes.get(j));
                    }
                } else {
                    done = true;
                    break;
                }
                i++;
            }
            if(done) {
                println("YES");
                println(list.size());
                for(int j=0;j<list.size();j++)
                    print(list.get(j) + " ");
                println("");
            } else {
                println("NO");
            }
        }
    }

    /************************ SOLUTION ENDS HERE ************************/



    /************************ TEMPLATE STARTS HERE **********************/

    public static void main(String[] args) throws IOException {
        reader = new BufferedReader(new InputStreamReader(System.in));
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false);
        st = null;
        solve();
        reader.close();
        writer.close();
    }

    static BufferedReader  reader;
    static PrintWriter     writer;
    static StringTokenizer st;

    static String next() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                String line = reader.readLine();
                if (line == null) {
                    return null;
                }
                st = new StringTokenizer(line);
            } catch (Exception e) {
                throw new RuntimeException();
            }
        }
        return st.nextToken();
    }

    static String nextLine() {
        String s = null;
        try {
            s = reader.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return s;
    }

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

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

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

    static char nextChar() {
        return next().charAt(0);
    }

    static int[] nextIntArray(int n) {
        int[] a = new int[n];
        int i = 0;
        while (i < n) {
            a[i++] = nextInt();
        }
        return a;
    }

    static long[] nextLongArray(int n) {
        long[] a = new long[n];
        int i = 0;
        while (i < n) {
            a[i++] = nextLong();
        }
        return a;
    }

    static int[] nextIntArrayOneBased(int n) {
        int[] a = new int[n + 1];
        int i = 1;
        while (i <= n) {
            a[i++] = nextInt();
        }
        return a;
    }

    static long[] nextLongArrayOneBased(int n) {
        long[] a = new long[n + 1];
        int i = 1;
        while (i <= n) {
            a[i++] = nextLong();
        }
        return a;
    }

    static void print(Object o) {
        writer.print(o);
    }

    static void println(Object o) {
        writer.println(o);
    }

    static void printArr(int[] arr) {
        for(int i = 0 ; i < arr.length; i++)
            System.out.print(arr[i] + " ");
        System.out.println();
    }

    /************************ TEMPLATE ENDS HERE ************************/
}