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>
//#include<unordered_set>
     #define ll long long int
    #define fi first
    #define se second
    #define m_p make_pair
    #define endl '\n'
    #define fast_io ios_base::sync_with_stdio(0); cin.tie(0)
    #define endl '\n'
    #define loop(i, a, b) for (long long unsigned int i = a; i < b; i++)
    #define pb push_back
    #define all(v) v.begin(),v.end()
    using namespace std;

   // typedef long long ll;
     typedef pair<ll,ll> pairs;
    const int MAXN = 412345;
    const int MAXINT = 2047483098;
    const ll MOD = 1e9 + 7;
    const int MAX = 1e4;
     const long double pi=2*acosl(0);
    const long double EPS = 1e-10;
     int gcd(int a,int b){if(b==0)return a;return gcd(b,a%b);}

bool isPrime(long long unsigned n){if(n<=1)return false;if(n<=3)return true;
	if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)
	if(n%i==0||n%(i+2)==0)return false;return true;}

int nCr(int n,int r){int res=1;if(n<r)return -1;r=min(r,n-r);
	loop(i,0,r){res*=(n-i);res/=(i+1);}return res;}

int nPr(int n,int r){int res=1;if(n<r)return -1;
	loop(i,0,r)res*=(n-i);return res;}
   bool comp(ll &a,ll &b){
       return a>b;
   }
void solve(){
 ll n;
 cin>>n;
 vector<ll> vec(n);
 map<ll,bool> mp;
 loop(i,0,n){
     cin>>vec[i];
     mp[vec[i]]=true;
 }
ll s=n;
for(ll i=0;i<vec.size();i++){
   for(ll j=0;j<vec.size();j++){
       if(j==i)continue;
       ll a=abs(vec[i]-vec[j]);
       if(!mp[a]){
        s++;
        vec.pb(a);
        mp[a]=true;
       }
       if(s>300){
           cout<<"NO";
           return;
       }
   }
}
  if(s>300){
           cout<<"NO";
           return;
       }else{
           cout<<"YES"<<endl;
           cout<<vec.size()<<endl;
           loop(i,0,vec.size()){
               cout<<vec[i]<<" ";
           }
       }

}
        


    int main()
    {
        fast_io;

        int T=1 ;
      cin >> T;
      
        while(T--) {
            solve();
            cout<<endl;
        }

        return 0;
}