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"
#define faster ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define ll long long int
#define pb push_back
#define f(a) for(ll i = 0;i<a;i++)
#define ffor(i,a,b) for(ll i = a;i<b;i++)
#define rfor(i,a,b) for(ll i = a;i>b;i--)
#define sll set<ll>
#define mll map<ll,ll>
#define vll vector<ll>
#define pll pair<ll,ll>
#define mcl map<char,ll>
#define usll unordered_set<ll,ll>
#define umll unordered_map<ll,ll>
#define umcl unordered_map<char,ll>
#define pqg priority_queue<ll>
#define pqs priority_queue<ll,vll,greater<ll> >
#define T ll t; cin>>t; while(t--)
#define freeopen freopen("input.txt","r",stdin);
#define freeclose freopen("output.txt","w",stdout);
#define MOD 1000000007
#define setbits(x) __builtin_popcountll(x)
#define INF 1000000000000000000
#define all(x) x.begin(), x.end()
using namespace std;
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }
ll modpower(ll x,ll y,ll m){if(y==0)return 1;ll p=modpower(x,y/2,m)%m;p=(p*p)%m;return (y%2==0)?p:(x*p)%m;}


ll power(ll a,ll b)
{
    ll ans = 1;
    while(b) {
        if(b&1) ans = ans * a;
        a = a * a;
        b >>= 1;
    }
    return ans;
}

bool isPrime(ll n)
{

    if (n <= 1)
        return false;
    if (n <= 3)
        return true;

    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (ll i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}


void solve()
{
    
}

int main()
{
    faster
    T
    {
        ll n;
        cin>>n;
        ll a[n];
        vll v;
        map<ll,ll> m;
        f(n){
            cin>>a[i];
            v.pb(a[i]);
            m[a[i]]++;
        }

            ll si=v.size();
            // cout<<v.size()<<"\n";
            ll tf=1;
            for(ll i=0;i<si;i++)
            {
                
                for(ll j=0;j<si;j++)
                {
                    if(i!=j)
                    {
                        // m[(ll)abs(v[i]-v[j])]++;
                        if(m[(ll)abs(v[i]-v[j])]==0)
                        {
                            if(v.size()>300 || abs(v[i]-v[j])>1000000000)
                            {
                                tf=0;
                                break;
                            }
                            // break;

                            v.pb(abs(v[i]-v[j]));
                            m[(ll)abs(v[i]-v[j])]++;
                            si=v.size();
                        }
                    }
                }
            }
        
            
    
        // cout<<v.size()<<"\n";
        if(tf)
        {
            cout<<"yes\n"<<v.size()<<"\n";

            for(ll i=0;i<v.size();i++)
                cout<<v[i]<<" ";

            cout<<"\n";
        }
        else
            cout<<"no\n";
    }
}