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.util.*;

import java.lang.*;
import java.io.*;
import java.math.BigInteger;


public class CF {
	private static FS sc = new FS();
	
	private static class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());				               
	            } catch (IOException e) {}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	
	private static class extra {
		
		static int[] intArr(int size) {
			int[] a = new int[size];
			for(int i = 0; i < size; i++) a[i] = sc.nextInt();
			return a;
		}
		
		static long[] longArr(int size) {
			long[] a = new long[size];
			for(int i = 0; i < size; i++) a[i] = sc.nextLong();
			return a;
		}
		
		static long intSum(int[] a) {
			long sum = 0; 
			for(int i = 0; i < a.length; i++) {
				sum += a[i];
			}
			return sum;
		}
		
		static long longSum(long[] a) {
			long sum = 0; 
			for(int i = 0; i < a.length; i++) {
				sum += a[i];
			}
			return sum;
		}
		
		static LinkedList[] graphD(int vertices, int edges) {
			LinkedList[] temp = new LinkedList[vertices+1];
			for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
			for(int i = 0; i < edges; i++) {
				int x = sc.nextInt();
				int y = sc.nextInt();
				temp[x].add(y);
			}
			return temp;
		}
		
		static LinkedList[] graphUD(int vertices, int edges) {
			LinkedList[] temp = new LinkedList[vertices+1];
			for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
			for(int i = 0; i < edges; i++) {
				int x = sc.nextInt();
				int y = sc.nextInt();
				temp[x].add(y);
				temp[y].add(x);
			}
			return temp;
		}
		
		static void printG(LinkedList[] temp) {
			for(LinkedList<Integer> aa:temp) System.out.println(aa);
		}
		
	}
	
	static class sieve {
		static final int MAXN = (int) (Math.pow(10, 7)+5);
		
	    static int spf[] = new int[MAXN];
	    
	    static void sieve() {
	        spf[1] = 1;
	        for (int i=2; i<MAXN; i++) spf[i] = i;
	        for (int i=4; i<MAXN; i+=2) spf[i] = 2;
	        for (int i=3; i*i<MAXN; i++) if (spf[i] == i) for (int j=i*i; j<MAXN; j+=i) if (spf[j]==j) spf[j] = i;
	    }
	      
	    static long get(int x) {
	    	ArrayList<Integer> ret = new ArrayList<>();
	        while (x != 1) {
	            ret.add(spf[x]);
	            x = x / spf[x];
	        }
	        HashMap<Integer, Integer> temp = new HashMap<>();
	        for(int aa:ret) temp.put(aa, (temp.getOrDefault(aa, 0)+1)%2);
	        long pro = 1;
	        for(int aa:temp.keySet()) pro *= Math.pow(aa, temp.get(aa));
	        return pro;
	    }
	    
	}
	
	static LinkedList[] temp;
	static int mod = (int)Math.pow(10, 9) + 7;
	
	public static void main(String[] args) {
		int t = sc.nextInt();
//		int t = 1;
		StringBuilder ret = new StringBuilder();
		while(t-- > 0) {
			int n = sc.nextInt();
			int[] a = extra.intArr(n);
			int flag = 0;
			for(int aa:a) if(aa < 0) flag = 1;
			if(flag == 1) ret.append("NO\n");
			else {
				HashSet<Integer> temp = new HashSet<>();
				for(int aa:a) temp.add(aa);
				int j = 301;
				while(true) {
					HashSet<Integer> cur = new HashSet<>();
					cur.addAll(temp);
					for(int aa:cur) {
						for(int bb:cur) if(aa != bb) {
							temp.add(Math.abs(aa-bb));
						}
					}
					if(temp.size() == cur.size()) break;
					temp.addAll(cur);
				}
				ret.append("YES" + "\n" + temp.size() + "\n");
				for(int aa:temp) ret.append(aa + " ");
				ret.append("\n");
			}
		}
		System.out.println(ret);
	}
	
	
	
}