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 namespace std;
#define ll long long
#define pb push_back
#define mp make_pair
#define inf 100000000000
#define mod 1000000007
 double power(double x,ll n)
{
    if(n==0)
        return 1;
    else if(n%2 == 0)        //n is even
        return power((x*x),n/2);
    else                             //n is odd
        return (x*power((x*x),(n-1)/2));

}
ll C(ll n, ll r) {
    if(r > n - r) r = n - r; // because C(n, r) == C(n, n - r)
    long long ans = 1;
    int i;

    for(i = 1; i <= r; i++) {
        ans =((ans) * (n - r + i));
        ans =(ans)/i;
    }

    return ans;
}
ll modularExponentiation(ll x,ll n,ll M)
{
    if(n==0)
        return 1;
    else if(n%2 == 0)        //n is even
        return modularExponentiation((x*x)%M,n/2,M);
    else                             //n is odd
        return (x%M*modularExponentiation((x*x)%M,(n-1)/2,M)%M)%M;

}

ll modInverse(ll A,ll M)
{
    return modularExponentiation(A,M-2,M);
}

void solve()
{
    ll n,i,temp,temp1,diff;
    map<ll,ll> m1;
    queue<ll> q1;
    cin>>n;
    for(i=0;i<n;i++)
    {
        cin>>temp;
        m1[temp]++;
        q1.push(temp);
    }
    while(q1.size()!=0)
    {
        temp=q1.front();
        q1.pop();
        for(auto x:m1)
        {
            temp1=x.first;
            if(temp>temp1)
            {
                diff=temp-temp1;
                if(m1[diff]==0)
                {
                    m1[diff]=1;
                    q1.push(diff);
                }
            }
            else if(temp<temp1)
            {
                diff=temp1-temp;
                if(m1[diff]==0)
                {
                    m1[diff]=1;
                    q1.push(diff);
                }
            }
          if(m1.size()>300)
          {
              cout<<"NO"<<endl;
              return ;
          }
        }
    }
    cout<<"YES"<<endl;
    cout<<m1.size()<<endl;
    for(auto x:m1)
        cout<<x.first<<" ";
    cout<<endl;
}
int main()
{
    ll t;
    cin>>t;
    while(t--)
    solve();
   return 0;
}