description: 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.
solution: //package Java;
  import java.io.*;
  import java.text.DecimalFormat;
  import java.util.*;
  import java.util.Map.Entry;
  
  public class Main {
  	static final PrintWriter out = new PrintWriter(System.out, true);
      static final FastScanner sc = new FastScanner();
  	public static void main(String[] args)  {
  		int t=sc.nextInt();
  		while(t-->0) {
  			int n=sc.nextInt();
  			int a[]=sc.nextIntArray(n);int sum=0,in=-1;
  			for(int i=0;i<n;i++){
  				sum+=a[i];
  			}
  			if(isPrime(sum)==false) {
  				out.println(n);for(int i=0;i<n;i++) {
  					out.print((i+1)+" ");
  				}
  				out.println();
  			}else {
  				for(int i=0;i<n;i++) {
  					if(isPrime(sum-a[i])==false) {
  						in=i;
  						break;
  					}
  				}
  				out.println(n-1);
  				for(int i=0;i<n;i++) {
  					if(i!=in) out.print((i+1)+" ");
  				}
  				out.println();
  			}
  		}
  		
  		
  		
  		out.close();
  	}
  static long Cr(int n, int r,int mod,long fac[],long inv[]) {
      	if(n<r) return 0L;
  		return fac[n]*(inv[r]*inv[n - r]%mod)%mod;
   
  //    	long fac[]=new long[100000];
  //    	long inv[]=new long[100000];
  //    	fac[0]=fac[1]=1;
  //    	fac[i]=modmul(fac[i-1],i,m);
  //		inv[i] =modpow(fac[i], m - 2,m);
  //    	long ans = 1;
  //	    ans *= (Cr(n-x,more,m,fac,inv)*fac[more])%m;
  //	    ans %= m;
  }
  	static boolean [] sieveOfEratosthenes(int n){
  	    boolean prime[] = new boolean[n + 1];
  	    for (int i = 0; i <= n; i++)
  	        prime[i] = true;
  	
  	    for (int p = 2; p * p <= n; p++)
  	    {
  	        if (prime[p] == true)
  	        {
  	            for (int i = p * p; i <= n; i += p)
  	                prime[i] = false;
  	        }
  	    }
  	    prime[1]=prime[0]=false;
  	    return prime;
  	}
  	static void divisors(int n,ArrayList a[]) {
  		for(int i=1;i<=n;i++) {
  			for(int j=i;j<=n;j+=i) {
  				a[j].add(i);
  			}
  		}
  	}
  	 static long modDivide(long a, long b, int m)
  	  {
  	      a = a % m;
  	      long inv = modInverse(b, m);
  	      if (inv == -1) return -1;
  	      else
  	      {
  	        long c = (inv * a) % m ;
  	        return c;
  	      }
  	  }
  	  static long modInverse(long a, int m){
  	        long g = gcd(a, m);
  	        if (g != 1) return -1;
  	        else
  	        {
  	        	return modpow(a,m-2,m);
  	        }
  	    }
  	  public static double logx(long m,int x){
  	        double result = (double)(Math.log(m) / Math.log(x));
  	  
  	        return result;
  	    }
  	     static double setPrecision(double ans,int k){
  	    	 double y=Math.pow(10,k);
  	    	 return Math.round(ans*y)/y;
  	     }
  	  public static double log2(double m)
  	    {
  	        double result = (double)(Math.log(m) / Math.log(2));
  	  
  	        return result;
  	    }
  	  static int countDivisors(int n)
  	    { int ans=0;
  	        for (int i=1; i<=Math.sqrt(n); i++)
  	        {
  	            if (n%i==0)
  	            {
  	                if (n/i == i) ans++;
  //	                    System.out.print(" "+ i);
  	      
  	                else ans+=2;
  //	                    System.out.print(i+" " + n/i + " " );
  	            }
  	        }
  			return ans;
  	    }
  	  static ArrayList<Integer> pyyy = new ArrayList<Integer>();
  		static void sieveppp(int MAX) 
  		{ 
  		    int[] isPrime=new int[MAX+1]; 
  		  
  		    for (int i = 2; i<= MAX; i++) 
  		    { 
  		        if (isPrime[i] == 0) 
  		        { 
  		        	pyyy.add(i); 
  		            for (int j = 2; i * j<= MAX; j++) 
  		                isPrime[i * j]= 1; 
  		        } 
  		    } 
  		} 
  		 static int phi(int n) { 
  			 ///   call sieveppp(n) in main fn
  		      int res = n; 
  		      for (int i=0; pyyy.get(i)*pyyy.get(i) <= n; i++) 
  		      { 
  		          if (n % pyyy.get(i)== 0) 
  		          { 
  		              res -= (res / pyyy.get(i)); 
  		              while (n % pyyy.get(i)== 0) 
  		              n /= pyyy.get(i); 
  		          } 
  		      } 
  		      if (n > 1) 
  		      res -= (res / n); 
  		    
  		      return res; 
  		  } 
  	public static void sort(long[] arr){//because Arrays.sort() uses quicksort which is in worst O(n^2)
  		//Collections.sort() uses merge sort
  		ArrayList<Long> ttttt = new ArrayList<Long>();
  		for(long y: arr)ttttt.add(y);Collections.sort(ttttt);
  		for(int i=0; i < arr.length; i++)arr[i] = ttttt.get(i);
  	}
  	public static void sort(int[] arr){//because Arrays.sort() uses quicksort which is dumb
  		//Collections.sort() uses merge sort
  		ArrayList<Integer> ttttt = new ArrayList<>();
  		for(int y: arr)ttttt.add(y);Collections.sort(ttttt);
  		for(int i=0; i < arr.length; i++)arr[i] = ttttt.get(i);
  	}
  	static boolean check(double[] a,double m,int n,int k) {
  		int ans=0;
  		for(int i=0;i<n;i++) {
  			ans+=(Math.floor(a[i]/m));
  			if(ans>=k) return true;
  		}
  		return ans>=k;
  	}
  	     static boolean isKthBitSet(long n, int k) { 
  			if ((n & (1 << (k - 1))) > 0) 
  			return true;
  			else
  			return false;
  			
  		 } 
  	     static long modpow(long a,long b,int m) {
  	    	 if(b==0) return 1;
  	    	 if(b==1) return a;
  	    	 long ans=0;
  	    	 long t=modpow(a,b/2,m);
  	    	 ans=modmul(t,t,m);
  	    	 if(b%2!=0) ans=modmul(a,ans,m);
  	    	 return ans;
  	     }
  		static long modmul(long a,long b,int m) {
  			return mod(mod(a,m)*mod(b,m),m);
  		}
  		static long mod(long x,int m) {
  			return ((x%m)+m)%m;
  		}
  		 static int octaltodecimal(int deciNum)
  		    {
  		        int octalNum = 0, countval = 1;
  		        int dNo = deciNum;
  		        while (deciNum != 0) {
  		 
  		            // decimals remainder is calculated
  		            int remainder = deciNum % 8;
  		 
  		            // storing the octalvalue
  		            octalNum += remainder * countval;
  		 
  		            // storing exponential value
  		            countval = countval * 10;
  		            deciNum /= 8;
  		        }
  		       return octalNum;
  		    }
  		
  	  public static int floorSqrt(int x) 
  	    { 
  	        // Base Cases 
  	        if (x == 0 || x == 1) 
  	            return x; 
  	  
  	        // Do Binary Search for floor(sqrt(x)) 
  	        int start = 1, end = x, ans=0; 
  	        while (start <= end) 
  	        { 
  	            int mid = (start + end) / 2; 
  	  
  	            // If x is a perfect square 
  	            if (mid*mid == x) 
  	                return mid; 
  	  
  	            // Since we need floor, we update answer when mid*mid is 
  	            // smaller than x, and move closer to sqrt(x) 
  	            if (mid*mid < x) 
  	            { 
  	                start = mid + 1; 
  	                ans = mid; 
  	            } 
  	            else   // If mid*mid is greater than x 
  	                end = mid-1; 
  	        } 
  	        return ans; 
  	    } 
  		static int div(int n,int b) {
  			int g=-1;
  			for(int i=2;i<=Math.sqrt(n);i++) {
  				if(n%i==0&&i!=b) {
  					g=i;
  					break;
  				}
  			}
  			return g;
  		}
  	 
  	  
  	  public static long gcd(long a, long a2) 
  	    { 
  	        if (a == 0) 
  	            return a2; 
  	          
  	        return gcd(a2%a, a); 
  	    } 
  	   public static long lcm(long a, long b)  
  	  {  
  	     return (a*b)/gcd(a, b);  
  	  }  
         public static boolean isPrime(int n) {
             if (n <= 1) {
                 return false;
             }
             if (n == 2) {
                 return true;
             }
             if (n % 2 == 0) {
                 return false;
             }
             for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2) {
                 if (n % i == 0) {
                     return false;
                 }
             }
             return true;
         }
  	      
         static class FastScanner {
             BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
             StringTokenizer strToken = new StringTokenizer("");
             String next() {
          	   while (!strToken.hasMoreTokens()) {
          		   try {
          			   strToken = new StringTokenizer(input.readLine()); 
          		   }  catch (IOException e) { 
          			    e.printStackTrace();
          			  } 
          	   } return strToken.nextToken(); 
          	}
             String nextLine() {
          	   String strrrrr = ""; try {
          		   strrrrr = input.readLine(); 
          	   } catch (IOException e) {
          			   e.printStackTrace(); 
          		}
          	   return strrrrr; 
             }
             int nextInt() {
          	   return Integer.parseInt(next()); 
          	}
             int[] nextIntArray(int n) {
          	   int[] a = new int[n]; 
          	   for (int i = 0; i < n; i++) a[i] = nextInt(); 
          	   return a; 
             }
             long nextLong() {
          	   return Long.parseLong(next()); 
          	}
             long[] nextLongArray(int n) {
          	   long[] a = new long[n];
          	   for (int i = 0; i < n; i++) a[i] = nextLong(); 
          	   return a; 
          	}
             double nextDouble() {
          	   return Double.parseDouble(next()); 
             }
         }
  }
  class DisjointSet {
      
      private int[] parent;
      private int[] rank;
      
      public DisjointSet(int n) {
          if (n < 0) throw new IllegalArgumentException();
          parent = new int[n];
          for(int i=0;i<n;i++) parent[i]=i;
          rank = new int[n];
      }
      
      public int find(int x) {
          if (parent[x] == x) return x;
          return parent[x] = find(parent[x]); // Path compression by halving.
      }
      
      // Return false if x, y are connected.
      public boolean union(int x, int y) {
          int rootX = find(x);
          int rootY = find(y);
          if (rootX == rootY) return false;
          
          // Make root of smaller rank point to root of larger rank.
          if (rank[rootX] < rank[rootY]) parent[rootX] = rootY;
          else if (rank[rootX] > rank[rootY]) parent[rootY] = rootX;
          else {
              parent[rootX] = rootY;
              rank[rootY]++;
          }
          
          return true;
      }
  }
  
  
  class Pair{
    int value;
    int distance;
    Pair(int value,int distance){
        this.value=value;
        this.distance=distance;
    }
    
  }
  class sortbydistance implements Comparator<Pair>{
  
  	@Override
  	public int compare(Pair o1, Pair o2) {
  		return o1.distance-o2.distance;
  	}
  	
  }
  class Solution
  {
    static int[] dijkstra(int n, ArrayList<ArrayList<ArrayList<Integer>>> adj, int S)
    {
    	 PriorityQueue<Pair>set=new PriorityQueue<>(new sortbydistance());
        int dis[]=new int[n];
        Arrays.fill(dis,Integer.MAX_VALUE);
        dis[0]=0;
        while(set.isEmpty()==false) {
        	Pair p=set.poll();
        	int u=p.value;
        	for(int i=0;i<adj.get(u).size();i++) {
        		int v=adj.get(u).get(i).get(0);
        		int wt=adj.get(u).get(i).get(1);
        		if(dis[v]>dis[u]+wt) {
        			if(dis[v]!=Integer.MAX_VALUE) {
        				set.remove(new Pair(dis[v],v));
        			}
        			dis[v]=dis[u]+wt;
        			set.add(new Pair(v,dis[v]));
        		}
        		
        	}
        }
        
        
        return dis;
    }
  }
  
  
  /*    Points- 
   * 
   *    1) when need max and min both use treeset if duplicates are there then use treeset<int []> and add a[1] as unique
   *    number but imp is to make new TreeSet<>((x,y)->(x[0]==y[0])?(x[1]-y[1]):(x[0]-y[0]))  in this x[0]==y[0] cond.
   *    is important 
   *    
   *    2) n!/2 is not fac[n]/2 it is modmul(fac[n],inv[2],m)
   *    
   *    3) if sorting can think of TreeSet also bro or want max and min use tree set as it has last and first function
   *    
   *    4) if dp is not optimisable then think of changing the defination of dp states
  */
  
  class pairs{
  	int a,b;
  	pairs(int a,int b){
  		this.a=a;this.b=b;
  	}
  }
  