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>
using namespace std;

#define ll long long
#define inf 1000000000000000000
#define f(i,a,b) for(int i=a;i<b;i++)
#define mod 1000000007
#define yes cout<<"YES\n";
#define no cout<<"NO\n";
#define npr(n,r) (fact[n]/fact[n-r])
#define ncr(n,r) (fact[n]/(fact[n-r]*fact[r]))
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) ((a*b)/__gcd(a,b))


int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    /*while(t--){
        int n,flag=0,pp=-1,lol=0;
        ll hcf;
        cin>>n;
        ll a[n];
        f(i,0,n){
            cin>>a[i];
            if(a[i]<0){flag=1;}
            else if(a[i]==0){pp=i;flag=2;}
        }
        long long int*q=std::max_element(a,a+n);
        if(flag==1)no
        else{yes
            if(flag==2 and n==2){cout<<2<<endl<<a[0]<<" "<<a[1]<<endl;}

            else if(pp==-1){
                hcf=gcd(a[0],a[1]);
                f(i,2,n){
                    hcf=gcd(a[i],hcf);
                    if(hcf==1){lol=1;
                        cout<<a[*q]<<endl;
                        f(j,1,a[*q]+1){cout<<j<<" ";}
                        cout<<endl;
                        break;}
                }
                if(lol==0){
                    cout<<*q/hcf<<endl;
                    for(int i=hcf;i<=*q;i+=hcf)cout<<i<<" ";
                    cout<<endl;
                }
            }

            else if(pp==0){
            ll hcf=gcd(a[1],a[2]);
            f(i,3,n){
                hcf=gcd(a[i],hcf);
                if(hcf==1){lol=1;cout<<a[*q]<<endl;
                    f(j,0,a[*q]+1){cout<<j<<" ";}
                    cout<<endl;
                    break;}
                }
                if(lol==0){
                    cout<<*q/hcf+1<<endl;
                    for(int i=0;i<=*q;i+=hcf)cout<<i<<" ";
                    cout<<endl;
                }
            }

            else if(pp==1){
            ll hcf=gcd(a[0],a[2]);
            f(i,3,n){
                hcf=gcd(a[i],hcf);
                if(hcf==1){lol=1;cout<<a[*q]<<endl;
                    f(j,0,a[*q]+1){cout<<j<<" ";}
                    cout<<endl;
                    break;}
                }
                if(lol==0){
                    cout<<*q/hcf+1<<endl;
                    for(int i=0;i<=*q;i+=hcf)cout<<i<<" ";
                    cout<<endl;
                }
            }

            else{
                hcf=gcd(a[0],a[1]);
                f(i,2,n){
                    if(pp==i)continue;
                    hcf=gcd(a[i],hcf);
                    if(hcf==1){lol=1;
                        cout<<a[*q]<<endl;
                        f(j,0,a[*q]+1){cout<<j<<" ";}
                        cout<<endl;
                        break;}
                }
                if(lol==0){
                    cout<<*q/hcf+1<<endl;
                    for(int i=0;i<=*q;i+=hcf)cout<<i<<" ";
                    cout<<endl;
                }
            }

        }


    }*/
    while(t--){
        int n,flag=0;
        cin>>n;
        int a[n];
        f(i,0,n){cin>>a[i];if(a[i]<0){flag=1;cout<<"NO"; }}
        if(flag==0){yes cout<<101<<endl; f(i,0,101)cout<<i<<" ";}
        cout<<endl;
    }
}