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 CPP):

Solution:

#include"bits/stdc++.h"
#include<string>
using namespace std;

long long gcd(long long a, long long b)
{ // returns hcf of 2 numbers
	if(a==0 || b==0 || a==b) return a==b ? a : (a>0 ? a : b);
    return a>b ? gcd(b,a%b) : gcd(a,b%a);
    // or GCD(x,y)=GCD(x-y,y)
}
long long int po(int a,long long int n)
{
	int sum=1;
	for(long long int i=1;i<=n;i++)
	{
		sum=sum*a;
	}
	return sum;
}
int isSorted(int a[],int n){
		for(int i=1;i<n;i++){
		if(a[i-1]>a[i]){
			return 0;
		}
	}
		return 1;
	}
	
bool isSort(vector<int>s){
	vector<int>a=s;
	sort(a.begin(),a.end());
	if(s==a)return 1;
	else return 0;
}

bool isPrime(int n){
    
    if (n <= 1)
        return false;
    for (int i = 2; i < n; i++){
        if (n % i == 0)
            return false;
        }
  
    return true;
}


long long int eleven( int c){
long long int m=0;
while(c){
	m=m*10+1;
	c--;
}
	return m;
}
int main()
{
	ios_base::sync_with_stdio(NULL);  cin.tie(0); cout.tie(0); //thank you
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

int t;
cin>>t;
while(t--){
long long int n;
cin>>n;
vector<long long int>v;
bool b=1;
for(int i=0;i<n;i++){
	long long int input;
	cin>>input;
	
	if(input>=0)v.push_back(input);
	else b=0;
}


if(b==1){
	cout<<"YES"<<endl;
//	long long int y=gcd(v[0],v[1]);
//	for(int i=0;i<n;i++){
//	for(int j=i+1;j<n;j++){
//		if(gcd(v[i],v[j])<y)y=gcd(v[i],v[j]);
//	}	
//	}
//	
	long long int r=*max_element(v.begin(),v.end());
//	long long int lim =r/y;
//	vector<int>q;
//	for(int i=0;i<=lim;i++){
//		q.push_back(i*y);
//	}
cout<<r+1<<endl;
for(int i=0;i<=r;i++){
		cout<<i<<" ";
	}
//	cout<<q.size()<<endl;
//	for(int i=0;i<q.size();i++){
//		cout<<q[i]<<" ";
//	}
	cout<<endl;
}
else cout<<"NO"<<endl;
}

	return 0;
}