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:
//go to line 54 for some useful code.
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "debug.hpp"
#else 
#define dbg(...) 47
#define dbgm(...) 47
#endif
// refer https://codeforces.com/blog/entry/66279
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
//refer-> https://codeforces.com/blog/entry/62393 
struct custom_hash {
   static uint64_t splitmix64(uint64_t x) {
       // http://xorshift.di.unimi.it/splitmix64.c
       x += 0x9e3779b97f4a7c15;
       x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
       x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
       return x ^ (x >> 31);
   }
   size_t operator()(uint64_t x) const {
       static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
       return splitmix64(x + FIXED_RANDOM);
   }
};
typedef long long ll;
typedef unsigned long int ul;
typedef unsigned long long int ull;
typedef unsigned int ui;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef vector<vector<ll>> matrix;
typedef vector<ll> vll;
typedef vector<int> vii;
typedef unordered_map<int,int, custom_hash> umap;
typedef unordered_set<int, custom_hash>uset;
#define f(i, x, n) for (int i = x; i < n; i++)
#define rf(i, n, x) for(int i=n;i>=x;--i)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define F first
#define S second
#define pb push_back
#define endl "\n"
#define unique(v) v.erase(unique(v.begin(), v.end()), v.end());
#define mem(a, b) memset(a, b, sizeof(a))
#define fast_io() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define oo INT64_MAX
#define rng (ll)(1e5+10)
#define mod (ll)(1e9+7)
// write your code here....
void solve(){
	ll n;
	cin>>n;
	vll a(n);
	for(auto &aa:a)cin>>aa;
	bool p = 0, neg = 0;
	f(i, 0, n){
		if(a[i]>0)p = 1;
		if(a[i]<0) neg= 1;
	}
	if(neg){
		cout<<"NO"<<endl;
		return;
	}
	cout<<"YES"<<endl;
	vii ans;
	f(i, 0, 101) ans.pb(i);
	cout<<ans.size()<<endl;
	for(auto &aa:ans)cout<<aa<<" ";
	cout<<endl;
}
int main()
{
	fast_io();
	cerr << "...............Console is yours! :)................." << endl;
	//freopen("input.in","r",stdin);
	int T;
	cin>>T;
	int O = T;
	while(T--){
		solve();
	}
	cerr<<".......^_^........."<<endl;
	return 0;
}