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 <algorithm>
#include <string>
#include <vector>
#include <map>
#include <ctype.h>
#include <cmath>
#include <cstdio>
#include <set>
#include <iterator>
#define pb push_back
#define lond long double
#define mp make_pair
#define all(a) a.begin(), a.end()
#define endl '\n'
typedef long long lon;
typedef unsigned long long ulon;

using namespace std;

bool camp(string a, string b){
    return a < b;
}

lon ctoi(char a){
    return a - '0';
}

vector<lon> del(lon x){
    lon square_root = sqrt(x) + 1;
    vector<lon> a;
    for (lon i = 1; i < square_root; i++) {
        if (x % i == 0 && i*i != x){
            a.pb(i);
            a.pb(x / i);
        }
        else if (x % i == 0 && i*i == x) a.pb(i);
    }
    return a;
}

int main(){
    ios_base::sync_with_stdio(false); cin.tie(0); cout.setf(ios::fixed); cout.precision(0); //freopen("ladder.in", "r", stdin);

    int t, n, x;
    cin >> t;
    while(t--){
        cin >> n;
        set<int> b;
        int f=0, mi=10000;
        for(int i=0;i<n;++i){cin >> x; b.emplace(x); mi=min(mi, x);}
        if(mi < 0){cout << "NO" << endl; continue;}
        while(b.size() != f && b.size() <= 300){
            f = b.size();
            for(auto p: b){
                for(auto y: b){
                    if(p==y) continue;
                    b.emplace(abs(p - y));
                }
            }
        }
        if(b.size() > 300){cout << "NO" << endl; continue;}
        cout << "YES" << endl;
        cout << b.size() << endl;
        for(auto p: b) cout << p << " ";
        cout << endl;
    }
}