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: import java.io.*;import java.lang.*;import java.util.*;
  //* --> number of prime numbers less then or equal to x are  -->  x/ln(x)
  //* --> String concatenation using the + operator within a loop should be avoided. Since the String object is immutable, each call for concatenation will
  // result in a new String object being created.
  // THE SIEVE USED HERE WILL RETURN A LIST CONTAINING ALL THE PRIME NUMBERS TILL N
   public class naive {static FastScanner sc;static PrintWriter pw;static class FastScanner {InputStreamReader is;BufferedReader br;StringTokenizer st;
  public FastScanner() {is = new InputStreamReader(System.in);br = new BufferedReader(is);}
  String next() throws Exception {while (st == null || !st.hasMoreElements())st = new StringTokenizer(br.readLine());
  return st.nextToken();}int nextInt() throws Exception {return Integer.parseInt(next());}long nextLong() throws Exception {
  return Long.parseLong(next());}int[] readArray(int num) throws Exception {int arr[]=new int[num];
  for(int i=0;i<num;i++)arr[i]=nextInt();return arr;}String nextLine() throws Exception {return br.readLine();
  }} public static boolean power_of_two(int a){if((a&(a-1))==0){ return true;}return false;}
  static boolean PS(double x){if (x >= 0) {double i= Math.sqrt(x);if(i%1!=0){
  return false;}return ((i * i) == x);}return false;}public static int[] ia(int n){int ar[]=new int[n];
  return ar;}public static long[] la(int n){long ar[]=new long[n];return ar;}
  public static void print(int ans,int t){System.out.println("Case"+" "+"#"+t+":"+" "+ans);}
  static long mod=1000000007;static int max=Integer.MIN_VALUE;static int min=Integer.MAX_VALUE;
  public static void sort(long[] arr){//because Arrays.sort() uses quicksort which is dumb
  //Collections.sort() uses merge sort
  ArrayList<Long> ls = new ArrayList<Long>();for(long x: arr)ls.add(x);Collections.sort(ls);
  for(int i=0; i < arr.length; i++)arr[i] = ls.get(i);}public static long fciel(long a, long b) {if (a == 0)  return 0;return (a - 1) / b + 1;}
   static boolean[] is_prime = new boolean[1000001];static HashSet<Integer> list = new HashSet<>();
  static long n = 1000000;public static void sieve() {Arrays.fill(is_prime, true);
  is_prime[0] = is_prime[1] = false;for (int i = 2; i * i <= n; i++) {
  if (is_prime[i]) {for (int j = i * i; j <= n; j += i)is_prime[j] = false;}}for (int i = 2; i <= n; i++) {
  if (is_prime[i]) {list.add(i);}}}
  
                          // ----------   NCR   ---------- \
  
  static int NC=100005;
  static long inv[]=new long[NC];
  static long fac_inv[]=new long[NC];
  static long fac[]=new long[NC];public static void initialize()
  {
      long MOD=mod;
     int  i;
      inv[1]=1;
      for(i=2;i<=NC-2;i++)
          inv[i]=(MOD-MOD/i)*inv[(int)MOD%i]%MOD; 
      fac[0]=fac[1]=1;
      for(i=2;i<=NC-2;i++)
          fac[i]=i*fac[i-1]%MOD;
      fac_inv[0]=fac_inv[1]=1;
      for(i=2;i<=NC-2;i++)
          fac_inv[i]=inv[i]*fac_inv[i-1]%MOD;
  }
  public static long  ncr(int  n,int  r)
  {
      long MOD=mod;
      if(n<r) return 0;
      return (fac[n]*fac_inv[r]%MOD)*fac_inv[n-r]%MOD;
  }
                          // ----------   NCR   ---------- \
                     
                         // ----------   FACTORS --------  \
  static int div[][] = new int[1000001][];
  public static void factors()
  {
    int divCnt[] = new int[1000001];
                         
    for(int i = 1000000; i >= 1; --i) {
    for(int j = i; j <= 1000000; j += i)
    divCnt[j]++;
    }
                   
                     
    for(int i = 1; i <= 1000000; ++i)
    div[i] = new int[divCnt[i]];
                   
    int ptr[] = new int[1000001];
    for(int i = 1000000; i >= 1; --i) {
    for(int j = i; j <= 1000000; j += i)
    div[j][ptr[j]++] = i;
      }   
  }    
                      // ----------   FACTORS --------  \
                          
  public static void main(String args[]) throws java.lang.Exception {
  sc = new FastScanner();pw = new PrintWriter(System.out);StringBuilder s = new StringBuilder();
  sieve();
  int t=sc.nextInt();
  while(t-->0)
  {
      int n=sc.nextInt();
    int ar[]=new int [n];
    int sum=0;
      for(int i=0;i<n;i++)
      {
          ar[i]=sc.nextInt();
          sum+=ar[i];
      }    
     if(!list.contains(sum))
  {
      s.append(n+"\n");
      for(int i=1;i<=n;i++)
      {
          s.append(i+" ");
      }
  }
  else{
      for(int i=0;i<n;i++)
  {
      if(!list.contains(sum-ar[i]))
      {
          s.append(n-1+"\n");
          for(int j=1;j<=n;j++)
          {
              if(j==i+1)
              {
                  continue;
              }
              s.append(j+" ");
          }
          break;
      }
  }
  }
  if(t>0)
  {
      s.append("\n");
  }}
  pw.print(s);pw.close();}}
  