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 ll long long
#define nl '\n'
#define vpn 1000000
using namespace std;
ll t,n,dismin;
ll a[vpn],b[vpn];
ll gcd(ll x, ll y)
{
    if (y==0) return x;
    return gcd(y,x%y);
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin>>t;
    while (t--)
    {
        cin>>n;
        for(ll i=1;i<=n;i++)
            cin>>a[i];
        sort(a+1,a+n+1);
        if (a[1]<0) cout<<"NO"<<nl;
        else
        {
            if (a[n]<=300)
            {
                cout<<"YES"<<nl;
                if (a[1]!=0)
                {
                    cout<<a[n]<<nl;
                for (ll i=1;i<=a[n];i++)
                    cout<<i<<" ";
                }
                else
                {
                    cout<<a[n]+1<<nl;
                    for (ll i=0;i<=a[n];i++)
                        cout<<i<<" ";
                }
                cout<<nl;
            }
            else
            {
                bool ans=true;
                vector<ll> dis;
                for (ll i=2;i<=n;i++)
                    dis.push_back(a[i]-a[i-1]);
                ll pos;
                pos=dis[0];
                for (ll i=1;i<dis.size();i++)
                {
                    pos=gcd(pos,dis[i]);
                    if (pos==1)
                    {
                        cout<<"NO"<<nl;
                        ans=false;
                        break;
                    }
                }
                if (ans==true)
                {
                    if (((a[n]-pos)/pos+1)>300) cout<<"NO"<<nl;
                    else
                    {
                        cout<<"YES"<<nl;
                        if (a[1]==0)
                        {
                            cout<<((a[n]-pos)/pos+2)<<nl;
                            cout<<a[1]<<" ";
                        }
                        else cout<<((a[n]-pos)/pos+1)<<nl;
                        for (ll i=pos;i<=a[n];i+=pos)
                            cout<<i<<" ";
                        cout<<nl;
                    }
                }
            }
        }
    }
}