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 ll = long long int ;
typedef long double ld;
#define loop(i,a,b) for(ll i = a ; i<b ; i++)
#define N cout<<"\n";
#define vi vector<ll>
#define mii map<ll,ll>
#define vpi vector<pair<ll,ll>>
#define f first
#define s second
#define pb push_back
#define MOD 1000000007
#define MAX 1000001

#define test4(x,y,z,a) cout<<"x is "<<x<<"		y is "<<y<<"		z is "<<z<<"		a is "<<a<<endl;
#define test3(x,y,z) cout<<"x is "<<x<<"		y is "<<y<<"		z is "<<z<<endl;
#define test2(x,y) cout<<"x is "<<x<<"		y is "<<y<<endl;
#define test1(x) cout<<"x is "<<x<<endl;
using namespace std;
/// seive of primes number ////
vector<bool>seive(MAX , true);
vi primes;
void createseive(){
    for (ll i = 2 ; i<MAX ; i++){
        if (seive[i]){
            primes.pb(i);
            for (ll j = i*i ; j<MAX ; j += i){
                seive[j] = false;
            }
        }
    }
}
void printprimes(ll n){
    for (int i = 0 ; i<n ; i++){
        cout<<primes[i]<<" ";
    }
}
/* seive codes end here */

/* power of a number code */
ll power(ll base , ll n){
    ll ans = 1 ;
    while (n!= 0){
        if (n&1){
            ans = (ans*base)%MOD;
            n = n-1;
        }
        else{
            base = (base*base)%MOD;
            n = n >> 1;
        }
    }
    return ans;
}
/*power code ends here */

int main() {

    ll t;
    cin >> t;
    while (t--){
        set<ll>s;
        ll n ;
        cin >> n;
        bool ansflag = false;
        vector<ll>a(n , 0);
        for (auto &x : a){
            cin >> x;
        }
        vector<ll>d = a;
        while (1 && d.size()<300){

            for (auto x : d){
                s.insert(x);
            }
            ll inserted = 0;
            for (int i = 0 ; i<d.size() ; i++){
                for (int j = i+1 ; j<d.size() ; j++){
                    if (s.find(abs(d[i]-d[j])) == s.end()){
                        s.insert(abs(d[i]-d[j]));
                        inserted++;
                    }
                }
            }
            if (inserted == 0){
                ansflag = true;
                break;
            }
            else{
                d.clear();
                for (auto x : s){
                    d.push_back(x);
                }
            }
            s.clear();
        }
        if (ansflag) {
            cout << "yes" << "\n";
            cout << d.size() << "\n";
            for (auto x : d) {
                cout << x << " ";
            }
            cout << "\n";
        }
        else{
            cout<<"NO"<<"\n";
        }
    }
    return 0;
}