Problem: A championship is held in Berland, in which n players participate. The player with the number i has a_i (a_i ≥ 1) tokens.

The championship consists of n-1 games, which are played according to the following rules:

  * in each game, two random players with non-zero tokens are selected; 
  * the player with more tokens is considered the winner of the game (in case of a tie, the winner is chosen randomly); 
  * the winning player takes all of the loser's tokens; 



The last player with non-zero tokens is the winner of the championship.

All random decisions that are made during the championship are made equally probable and independently.

For example, if n=4, a = [1, 2, 4, 3], then one of the options for the game (there could be other options) is: 

  * during the first game, the first and fourth players were selected. The fourth player has more tokens, so he takes the first player's tokens. Now a = [0, 2, 4, 4]; 
  * during the second game, the fourth and third players were selected. They have the same number of tokens, but in a random way, the third player is the winner. Now a = [0, 2, 8, 0]; 
  * during the third game, the second and third players were selected. The third player has more tokens, so he takes the second player's tokens. Now a = [0, 0, 10, 0]; 
  * the third player is declared the winner of the championship. 



Championship winners will receive personalized prizes. Therefore, the judges want to know in advance which players have a chance of winning, i.e have a non-zero probability of winning the championship. You have been asked to find all such players. 

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The first line of each test case consists of one positive integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of players in the championship.

The second line of each test case contains n positive integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the number of tokens the players have.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5. 

Output

For each test case, print the number of players who have a nonzero probability of winning the championship. On the next line print the numbers of these players in increasing order. Players are numbered starting from one in the order in which they appear in the input. 

Example

Input


2
4
1 2 4 3
5
1 1 1 1 1


Output


3
2 3 4 
5
1 2 3 4 5 

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.*;
import java.util.*;

public class Main {
	private BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	private BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
	private StringTokenizer st;
	private Long MOD = 1000000007L;
	private final String endl = "\n";

	public static void main(String[] args) throws Exception {
		new Main().solve();
	}

	private void solve() throws Exception {

		int testCases = 1;
		testCases = Integer.parseInt(in.readLine().trim());
		preProcess();
		for (int i = 1; i <= testCases; i++) {
			solveTestCase(i);
		}

		out.flush();
		out.close();
	}

	void preProcess() {

	}

	private void solveTestCase(int testCaseNumber) throws Exception {
		int n = ri();
		long arr[] = rla(n);
		
		if (n == 1) {
			out.write("1\n");
			out.write("1\n");
			return;
		}
		long copy[] = new long[n];
		
		
		for (int i = 0; i < n; i++) {
			copy[i] = arr[i];
		}
		
		
		ruffleSort(arr);
		
		long pre[] = new long[n];
		
		pre[0] = arr[0];
		for (int i = 1; i < n; i++) {
			pre[i] = arr[i] + pre[i-1];
		}
		
		
		int i = n -2 ;
		for (; i >= 0; i--  ) {
			if (i == 0) {
				if (arr[0] < arr[1]) {
					break;
				}
			}else {
				if (pre[i] >= arr[i+1]) {
					continue;
				} else {
					break;
				}
			}
		}
		
		long limit = i == -1 ? Long.MIN_VALUE : arr[i];
		int ans = 0;
		for (i = 0; i < n; i++ ) {
			if (copy[i] > limit) {
				ans += 1;
			}
		}
		out.write(ans + endl);
		for (i = 0; i < n; i++ ) {
			if (copy[i] > limit) {
				out.write(i + 1 + " ");
			}
		}
		out.write(endl);
		
		// check for last element as well
	}
	
	static void ruffleSort(int[] a) {
		//ruffle
		int n=a.length;
		Random r=new Random();
		for (int i=0; i<a.length; i++) {
			int oi=r.nextInt(n), temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
		
		//then sort
		Arrays.sort(a);
	}
	
	static void ruffleSort(long[] a) {
		//ruffle
		int n=a.length;
		Random r=new Random();
		for (int i=0; i<a.length; i++) {
			int oi=r.nextInt(n);
			long temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
		
		//then sort
		Arrays.sort(a);
	}

	boolean[] sieveOfEratosthenes(int n) {
		// Create a boolean array "prime[0..n]" and initialize
		// all entries it as true. A value in prime[i] will
		// finally be false if i is Not a prime, else true.
		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] is not changed, then it is a prime
			if (prime[p] == true) {
				// Update all multiples of p
				for (int i = p * p; i <= n; i += p)
					prime[i] = false;
			}
		}

		return prime;
	}

	// utils
	public long modInv(long a, long MOD) {
		return new BigInteger(a + "").modPow(new BigInteger((MOD - 2) + ""), new BigInteger(MOD + "")).longValue();
	}

	public long modPro(long a, long b, long MOD) {
		return ((a % MOD) * (b % MOD)) % MOD;
	}

	private long factorialMod(int n, long MOD) {
		long pro = 1;

		for (long i = 1; i <= n; i++) {
			pro = pro * i;
			pro %= MOD;
		}

		return pro;
	}

	// utilities below

	// IO UTILS

	void log(Object... args) {
		boolean testing = true;
		testing = false;

		if (!testing)
			return;

		for (Object s : args) {
			System.out.print(s.toString() + " ");
		}
		System.out.println();
	}

	String line() throws IOException {
		return in.readLine().trim();
	}

	StringTokenizer tokens() throws IOException {
		return new StringTokenizer(line());
	}

	long l(String s) {
		return Long.parseLong(s);
	}

	long rl() throws IOException {
		return l(line());
	}

	long[] rla(int n) throws IOException {
		long[] arr = new long[n];
		st = tokens();

		for (int i = 0; i < n; i++)
			arr[i] = l(st.nextToken());

		return arr;
	}

	int i(String s) {
		return Integer.parseInt(s);
	}

	int ri() throws IOException {
		return i(line());
	}

	int[] ria(int n) throws IOException {
		int[] arr = new int[n];
		st = tokens();

		for (int i = 0; i < n; i++)
			arr[i] = i(st.nextToken());

		return arr;
	}

	String rs() throws IOException {
		return line();
	}

	String[] rsa(int n) throws IOException {
		String s[] = new String[n];

		st = tokens();

		for (int i = 0; i < n; i++)
			s[i] = st.nextToken();

		return s;
	}

	double d(String s) {
		return Double.parseDouble(s);
	}

	double rd() throws IOException {
		return d(line());
	}

	double[] rda(int n) throws IOException {
		double[] arr = new double[n];
		st = tokens();

		for (int i = 0; i < n; i++) {
			arr[i] = d(st.nextToken());
		}

		return arr;
	}
	// IO UTILS END

	// MATH UTILS

	long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}
	// MATH UTILS END
}