A bow adorned with nameless flowers that bears the earnest hopes of an equally nameless person.

You have obtained the elegant bow known as the Windblume Ode. Inscribed in the weapon is an array of n (n ≥ 3) positive distinct integers (i.e. different, no duplicates are allowed).

Find the largest subset (i.e. having the maximum number of elements) of this array such that its sum is a composite number. A positive integer x is called composite if there exists a positive integer y such that 1 < y < x and x is divisible by y.

If there are multiple subsets with this largest size with the composite sum, you can output any of them. It can be proven that under the constraints of the problem such a non-empty subset always exists.

Input

Each test consists of multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first line of each test case contains an integer n (3 ≤ n ≤ 100) — the length of the array.

The second line of each test case contains n distinct integers a_{1},a_{2},...,a_{n} (1 ≤ a_{i} ≤ 200) — the elements of the array.

Output

Each test case should have two lines of output.

The first line should contain a single integer x: the size of the largest subset with composite sum. The next line should contain x space separated integers representing the indices of the subset of the initial array.

Example

Input


4
3
8 1 2
4
6 9 4 2
9
1 2 3 4 5 6 7 8 9
3
200 199 198


Output


2
2 1
4
2 1 4 3
9
6 9 1 2 3 4 5 7 8
3
1 2 3 

Note

In the first test case, the subset \\{a_2, a_1\} has a sum of 9, which is a composite number. The only subset of size 3 has a prime sum equal to 11. Note that you could also have selected the subset \\{a_1, a_3\} with sum 8 + 2 = 10, which is composite as it's divisible by 2.

In the second test case, the sum of all elements equals to 21, which is a composite number. Here we simply take the whole array as our subset.
// AUTHOR: fireLUFFYY, NITR
// Bhagwan, please AC karwa dena

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#pragma comment(linker, "/stack:200000000")

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define int long long
#define ull unsigned long long
#define vi vector<int>
#define vvl vector<vector<int>>
#define deql deque<int>
#define prquel priority_queue<int>
#define loop(i,n) for(int i=0;i<n;i++)
#define loopn(i,a,b) for(int i=a;i<=b;i++)
#define rloop(i,n) for(int i=n;i>0;i--)
#define sortv(a) sort(a.begin(),a.end())
#define sortvr(a) sort(a.rbegin(),a.rend())
#define verase(a) a.erase(unique(a.begin(),a.end()),a.end()) 
#define all(a) a.begin(),a.end()
#define getv(v,n) for(int i=0;i<n;i++){int x; cin>>x; v.push_back(x);};
#define dbg(x) cerr<<#x<<" = "; _print(x);cerr<<"\n";
#define mp make_pair
#define pb push_back
#define pf push_front
#define fi first
#define se second 
#define endl "\n"
#define inf 2e18
using namespace std;
using namespace __gnu_pbds;

int MOD=1000000007;
int mod=998244353;
const int N=200050;
const int nn=10050;
int fact[N]; // array to store values of factorial
bool prime[nn];	//array to store precalculated primes till 10^6
void _print(int a){cerr<<a;}  void _print(char a){cerr<<a;}  void _print(bool a){cerr<<a;}  void _print(string a){cerr<<a;}
template<class T> void _print(vector<T> v1){ cerr<<"[ "; for(T i:v1){_print(i);cerr<<" ";}cerr<<"]";}
template<class T> void _print(set<T> s1){ cerr<<"[ "; for(T i:s1){_print(i);cerr<<" ";}cerr<<"]";}
template<class T> void _print(map<T,T> m1){for(auto i:m1){cerr<<"[ "; _print(i.first);cerr<<" : ";_print(i.second);cerr<<" ]";cerr<<endl;}}
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
template<class T> T gcd(T a,T b){if(!a||!b)return a|b; unsigned shift=__builtin_ctz(a|b); a>>=__builtin_ctz(a); do{b>>=__builtin_ctz(b); if(a>b)swap(a,b);b-=a;}while(b); return a<<shift;}
template<class T> T lcm(T a,T b){ unsigned val=max(a,b)/gcd(a,b);val*=min(a,b);return val;}
bool pow2(int x){if(x&&(!(x&(x-1)))) return true; else return false;}
int multiply(int x, int y){return (x * 1ll * y) % MOD;}
int power(int x, int y){int z = 1;while(y){if(y & 1) z = multiply(z, x);x = multiply(x, x);y >>= 1;}return z;}
int modInverse(int x){return power(x, MOD - 2);}
int divide(int x, int y){return multiply(x, modInverse(y));}
void cal_factorial(){fact[0] = 1;for(int i = 1; i < N; i++)fact[i] = multiply(fact[i - 1], i);}// function to calculate factorial upto N
void cal_primes(){memset(prime,true,sizeof(prime)); loopn(i,2,sqrt(nn)){ if(prime[i]==true){ for(int j=i*i;j<=nn;j+=i){prime[j]=false;}}}}
int nCr(int n, int k){return divide(fact[n], multiply(fact[k], fact[n - k]));}
vector<int>Intersection(vi &v1,vi &v2){vi v3;sort(all(v1));sort(all(v2));set_intersection(all(v1),all(v2),back_inserter(v3));return v3;}
vector<int>Union(vi &v1,vi&v2){vi v3;sort(all(v1));sort(all(v2));set_union(all(v1),all(v2),back_inserter(v3));return v3;}
void FASTIO(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);}
int dx[8]={0,1,0,-1,1,1,-1,-1}; //direction vectors
int dy[8]={1,0,-1,0,1,-1,-1,1}; //direction vectors


void solve(int t)
{
	cal_primes();
	int testcases=t;
	while(t--)
	{
		//cout<<"#"<<(testcases-t)<<" Test-Case: "<<endl;
		//cout<<"Case #"<<(testcases-t)<<": ";
		int n;cin>>n;
		vi v;getv(v,n);
		int sum=0;

		for(int i=0;i<n;i++)
			sum+=v[i];

		if(!prime[sum])
		{
			cout<<n<<endl;
			for(int i=0;i<n;i++)
				cout<<i+1<<" ";
			cout<<endl;
		}
		else
		{
			bool flag=false;
			for(int i=0;i<n;i++)
			{
				int summ=sum-v[i];
				if(!prime[summ])
				{
					cout<<n-1<<endl;
					for(int j=0;j<n;j++){
						if(j!=i)
							cout<<j+1<<" ";
					}
					flag=true;
					cout<<endl;
				}
				if(flag)
					break;
			}
		}
	}
}

main()
{
	auto start=chrono::system_clock::now();
	{
		#ifndef ONLINE_JUDGE
			freopen("input.txt","r",stdin);
			freopen("output.txt","w",stdout);
			freopen("error.txt","w",stderr);
		#endif
		FASTIO();	
		int t=1;
		cin>>t;
		solve(t);
	}
	auto end=chrono::system_clock::now();
	chrono::duration<double> elapsed=end-start;
//	cout<<" Time taken: "<<elapsed.count()<<" sec";
	return 0;
}