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 int long long
#define ll          long long
#define pb          push_back
#define    endl        '\n'
#define pii         pair<ll ,ll >
#define vi          vector<int >
#define vii         vector<pii>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define mod       1000000007
#define rep(i,a,b)    for(ll  i=a;i<b;i++)
#define rep0(a)     for(ll i=0;i<a;i++)
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
#define mp          make_pair
#define itr(a,it)          for(typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define double        long double
using namespace std;
#define N  100005
#define M 1e9
ll bpow(ll a, ll b){ 
    int ans=1;
    while(b>0)
    {
        if(b&1)
            ans = ans*a%mod;
        b=b/2;
        a=a*a%mod;
    }
    return ans%mod;
}    
void solve()
{
    int n;
    cin>>n;
    vector<int>a(n);
    map<int,int>m;

    rep0(n) {cin>>a[i]; m[a[i]]=1;}
    // sort(all(a));
    while(1)
    {
        int f=0;
        rep0(n)
        {
            rep(j,0,n)
            {
                if(i==j)
                    continue;

                int x=abs(a[i]-a[j]);
                if(m[x]) continue;
                f=1;
                if(n>=300) 
                {
                    cout<<"NO";
                    return;
                }
                a.pb(x);
                m[x]=1;
                n++;
            }
        }
        if(f==0) break;
    }
    cout<<"YES"<<endl;
    cout<<n<<endl;
    rep0(n) cout<<a[i]<<" ";
} 
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    // freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    int TESTS=1;
    cin>>TESTS;
    while(TESTS--)
    {
        solve();
       cout<<endl;
    }
    return 0;
}