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<iostream>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<vector>
#include<algorithm>
#include<unordered_map>
#include<map>
#include<math.h>
#include<list>
#include<climits>
#include<bitset>
#include<set>
#define ll long long
#define vi vector<int>
#define vll vector<long long>
#define vvi vector<vector<int>>
#define vvll vector<vector<long long>>
using namespace std;

template<typename t>
void printArray(vector<t> &a)
{
    for(auto g : a)
    {
        cout<<g<<" ";
    }
}

template<typename t>
bool inputArray(vector<t> &a, int n,unordered_map<ll,int>&dp)
{
    a.resize(n);
    for(int i =0 ;i<n;i++)
    {
        cin>>a[i];
        if(dp.count(a[i]))
        {
            return false;
        }
        dp[a[i]]+=1;
    }
    return true;
}

void solve()
{
    int n;
    cin>>n;
    vll a;
    unordered_map<ll,int> dp;
    if(!inputArray<ll>(a,n,dp))
    {
        cout<<"NO"<<endl;
        return;
    }

    for(int i = 0;i<a.size();i++)
    {
        for(int j = i+1;j<a.size();j++)
        {
            if(dp.count(abs(a[i]-a[j])))
            {
                continue;
            }
            else{
                a.push_back(abs(a[i]-a[j]));
                if(a.size()>300)
                {
                    cout<<"NO"<<endl;
                    return;
                }
                dp[abs(a[i]-a[j])]+=1;
            }
        }
    }
    for(int i = 0;i<a.size();i++)
    {
        for(int j = i+1;j<a.size();j++)
        {
            if(dp.count(abs(a[i]-a[j])))
            {
                continue;
            }
            else{
                a.push_back(abs(a[i]-a[j]));
                if(a.size()>300)
                {
                    cout<<"NO"<<endl;
                    return;
                }
                dp[abs(a[i]-a[j])]+=1;
            }
        }
    }
    if(a.size()>300)
    {
        cout<<"NO"<<endl;
        return;
    }

    cout<<"YES"<<endl;
    cout<<a.size()<<endl;
    printArray<ll>(a);
    cout<<endl;
    return;
    
}

int main()
{

    int t;
    cin>>t;
    while(t--)
    {
        solve();
    }
return 0;
}