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: // Problem: A. Windblume Ode
  // Contest: Codeforces - Codeforces Round #749 (Div. 1 + Div. 2, based on Technocup 2022 Elimination Round 1)
  // URL: https://codeforces.com/contest/1586/problem/A
  // Memory Limit: 256 MB
  // Time Limit: 1000 ms
  // 
  // Powered by CP Editor (https://cpeditor.org)
  
  import java.io.*;
  import java.util.*;
  
  import static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  
  
  /**
   * zxz 
   * problem link:
   **/
  public class Main {
      //put some data here
  
      /**
       * do some clean action before next testcase
       */
      private void clear() {
  
      }
  
  	boolean [] flag = new boolean[20001];;
  
      private void solve() throws Exception {
          rf();
          int n = ri();
          int [] data = readArr(n);
          int sum = 0;
          for(int i : data){
          	sum += i;
          }
          if(!flag[sum]){
          	addAns(n);
          	for(int i = 1; i < n; i++){
          		sb.append(i + " ");
          	}
          	sb.append(n + "\n");
          	return ;
          }
          addAns(n - 1);
          for(int i = 0; i < n; i++){
          	if(!flag[sum - data[i]]){
          		for(int j = 0; j < n; j++){
          			if(j != i)
  	        			sb.append(j + 1 + " ");
  	        	}
  	        	sb.append("\n");
  	        	return ;
          	}
          }
      }
  
      private void run() throws Exception {
      	for(int i = 0; i <= 20000; i++){
      		flag[i] = isPrime(i);
      	}
         rf();
         int T = ri();
          // int T = 1;
  
          while (T-- > 0) {
              solve();
              if (T != 0) {
                  clear();
              }
          }
          printAns();
      }
  
      public static void main(String[] args) throws Exception {
          new Main().run();
      }
  
      StringBuilder sb = new StringBuilder();
      BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
      StringTokenizer st;
  
  	private void addAns(int a){
  		sb.append(a + "\n");
  	}
  	
  	private void addAns(long a){
  		sb.append(a + "\n");
  	}
  	
  	private void addAns(String s){
  		sb.append(s + "\n");
  	}
  
      private void rf() throws IOException {
          st = new StringTokenizer(infile.readLine());
      }
  
  
      private int ri() {
          return Integer.parseInt(st.nextToken());
      }
  
      private long rl() {
          return Long.parseLong(st.nextToken());
      }
  
  	private char [] rs2c(){
  		return st.nextToken().toCharArray();
  	}
  	
  	private String rs(){
  		return st.nextToken();
  	}
  
      private void printAns() {
          System.out.println(sb);
      }
  
      private int[] readArr(int N) throws Exception {
          int[] arr = new int[N];
          rf();
          for (int i = 0; i < N; i++) {
              arr[i] = Integer.parseInt(st.nextToken());
          }
          return arr;
      }
  
      private void print(int[] arr) {
          //for debugging only
          for (int x : arr) {
              System.out.print(x + " ");
          }
          System.out.println();
      }
      
      private void print(long[] arr) {
          //for debugging only
          for (long x : arr) {
              System.out.print(x + " ");
          }
          System.out.println();
      }
  
      private long[] readArr2(int N) throws Exception {
          long[] arr = new long[N];
          st = new StringTokenizer(infile.readLine());
          for (int i = 0; i < N; i++) {
              arr[i] = Long.parseLong(st.nextToken());
          }
          return arr;
      }
  
      private boolean isPrime(long n) {
          if (n < 2) {
              return false;
          }
          if (n == 2 || n == 3) {
              return true;
          }
          if (n % 2 == 0 || n % 3 == 0) {
              return false;
          }
          long sqrtN = (long) Math.sqrt(n) + 1;
          for (long i = 6L; i <= sqrtN; i += 6) {
              if (n % (i - 1) == 0 || n % (i + 1) == 0) {
                  return false;
              }
          }
          return true;
      }
  
      private long gcd(long a, long b) {
          if (a > b) {
              a = (a + b) - (b = a);
          }
          if (a == 0L) {
              return b;
          }
          return gcd(b % a, a);
      }
      
      private int gcd(int a, int b) {
          if (a > b) {
              a = (a + b) - (b = a);
          }
          if (a == 0) {
              return b;
          }
          return gcd(b % a, a);
      }
      
      private double pow(double x, long N) {
          double ans = 1.0;
          // 贡献的初始值为 x
          double x_contribute = x;
          // 在对 N 进行二进制拆分的同时计算答案
          while (N > 0) {
              if (N % 2 == 1) {
                  // 如果 N 二进制表示的最低位为 1，那么需要计入贡献
                  ans *= x_contribute;
              }
              // 将贡献不断地平方
              x_contribute *= x_contribute;
              // 舍弃 N 二进制表示的最低位，这样我们每次只要判断最低位即可
              N /= 2;
          }
          return ans;
      }
      
      long pow_mod(long x, long i, long mod){
  	    if(i == 0){
  	        return 1;
  	    }
  	    long temp = pow_mod(x, i >> 1, mod);
  	    temp = temp * temp % mod;
  	    if((i & 1 )!= 0){
  	        temp = temp * x % mod;
  	    }
  	    return temp;
  	}
      
  
  }
  