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.
import java.util.*;
import java.io.*;
public class Main {
    static BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
    static BufferedWriter wr = new BufferedWriter(new OutputStreamWriter(System.out));
    static StringTokenizer tok;
    public static void main(String[] args) throws Exception {
        solution();
    }
    public static void solution() throws Exception {
        /*  https://codeforces.com/contest/1586/problem/A :: Windblumne Ode
        *  나는 Windblume Ode라고 불리는 활을 얻었다. 활에는 n(n>=3) 인 서로 다른 양의 정수 배열이 새겨져있다.
        *  해당 배열의 부분집합 중에서 합이 합성수(소수가 아닌 수)가 되는 배열 중 가장 많은 요소를 갖고있는 부분을 찾아야한다.
        *   양의 합성수는 다음과 같은 경우에 성립한다.
        * 양의 정수 y에서 1 < y < x 이며, y로 나누어떨어지는 x가 존재할 경우 x를 합성수라고 한다.
        * 만약 여러개의 합성수 부분집합이 존재한다면, 그 중 아무 값이나 출력해도 된다. 문제에는 항상 부분집합이 존재함을 보장한다.
        * */
        int TestCase = Integer.parseInt(rd.readLine());
        for(int TT=0;TT<TestCase;TT++) {
            tok = new StringTokenizer(rd.readLine());
            int n = Integer.parseInt(tok.nextToken());
            tok = new StringTokenizer(rd.readLine());
            int[] arr = new int[n];
            for(int i=0;i<n;i++) {
                arr[i] = Integer.parseInt(tok.nextToken());
            }
            // 입력 테스트
            // System.out.println(Arrays.toString(arr));

            // 2를 제외한 모든 짝수가 합성수임.
            // 1. 모든 수를 더한 수가 합성수인지를 확인함, 합성수이면 출력
            // 2. 합성수가 아니라면, (무조건 홀수임) => 홀수가 존재한다면 홀수 한개를 제외하고 출력
            // 3. 모두 짝수일수는 없음 (모두 짝수라면 합도 짝수여야함)

            int sum = 0;
            String res = "";
            for(int temp : arr) sum += temp;
            boolean checkIsCom = false;
            for(int i=2;i<= (int)Math.sqrt(sum);i++)  {
                if(sum % i == 0) {
                    checkIsCom = true;
                    break;
                }
            }
            if(!checkIsCom) {
                // 소수라면 홀수를 찾아서 결과에서 제외함.
                boolean check = false;
                for(int i=0;i<n;i++) {
                    if(arr[i] % 2 != 0 && !check) {
                        check = true;
                    } else res += (i+1)+" ";
                }
                wr.write((n-1)+"");
                wr.newLine();
                wr.write(res);
            } else {
                for(int i=0;i<n;i++) res += (i+1)+" ";
                wr.write(n+"");
                wr.newLine();
                wr.write(res);
            }
            wr.newLine();
        }
        wr.flush();
        wr.close();
    }
}