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
const ll int mod = (ll)1e9+7;
#define endl "\n"

//do use flushout in interactive problems

//sieve using bitset
#define N 51000000
unsigned int prime[N / 64];
#define gP(n) (prime[n>>6]&(1<<((n>>1)&31)))
#define rP(n) (prime[n>>6]&=~(1<<((n>>1)&31)))
void sieve()
{
    memset( prime, -1, sizeof( prime ) );

    unsigned int i;
    unsigned int sqrtN = ( unsigned int )sqrt( ( double )N ) + 1;
    for( i = 3; i < sqrtN; i += 2 ) if( gP( i ) )
    {
        unsigned int i2 = i + i;
        for( unsigned int j = i * i; j < N; j += i2 ) rP( j );
    }
}

//calculate GCD in log(min(a,b))
ll int gcd (ll int a,ll int b) {
    while (b) {
        a %= b;
        swap(a, b);
    }
    return a;
}

//calculate a^b in logb 
ll int binexp(ll int a,ll int b){
    int ans=1;
    while(b!=0){
        if(b&1){
            ans=ans*a;
            b=b-1;
        }else{
            a*=a;
            b=b/2;
        }
    }
    return ans;
}
//calculate modular exp in logb
long long binpowm(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}

ll int rollinghash(string s){
    ll int pr=1;
    ll int hash=0;
    for(int i=0;i<s.length();i++){
        hash=(hash+((s[i]-'a'+1)*pr)%mod)%mod;
        pr*=31;//any prime greater than 26 for generating any newer formulas too
    }
    return hash;
}
/*-----------------------------------------------------------------------------
------------------------------------------------------------------------------*/



int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int t;cin>>t;
    while(t--){
        
        int n;cin>>n;
        int a[n+5];
        bool flag=false;
        for(int i=0;i<n;i++){
            cin>>a[i];
            if(a[i]<0){
                flag=true;
            }
        }
        
        
        if(flag==true){
            cout<<"No"<<endl;
        }else{
            cout<<"Yes"<<endl;
            cout<<101<<endl;
            for(int i=0;i<101;i++){
                cout<<i<<" ";
            }
            cout<<endl;
        }
    }
	
	return 0;
}