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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;

// CFPS -> CodeForcesProblemSet
public final class CFPSOptimized {
	static FastReader fr = new FastReader();
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) {
		int t = fr.nextInt();
		OUTER:
		for (int tc = 0; tc < t; tc++) {
			int n = fr.nextInt();
			Tuple[] tupsCum = new Tuple[n];
			Tuple[] tups = new Tuple[n];
			for (int i = 0; i < n; i++) {
				tups[i] = new Tuple();
				tupsCum[i] = new Tuple();
				tups[i].val = fr.nextLong();
				tupsCum[i].val = tups[i].val;
				tups[i].idx = i;
				tupsCum[i].idx = i;
			}
			Arrays.sort(tups);
			Arrays.sort(tupsCum);
			
			boolean[] canWin = new boolean[n];
			canWin[tups[n - 1].idx] = true;
			// Calculating the cumulance.
			for (int i = 1; i < n; i++) {
				tupsCum[i].val += tupsCum[i - 1].val;
			}
			
			/*for (int i = 0; i < n - 1; i++) {
				if (tupsCum[i].val >= tups[i + 1].val) {
					canWin[tups[i].idx] = true;
				}
			}*/
			
			for (int i = n - 2; i > -1; i--) {
				if (tupsCum[i].val >= tups[i + 1].val) {
					canWin[tups[i].idx] = canWin[tups[i + 1].idx] == true ? true : false;
				}
			}
			
			int winCtr = 0;
			for (int i = 0; i < n; i++)
				if (canWin[i])
					winCtr++;
			
			out.println(winCtr);
			
			for (int i = 0; i < n; i++)
				if (canWin[i])
					out.print((i + 1) + " ");
			
			out.println();
		}
		out.close();
	}
	
	static class Tuple implements Comparable<Tuple> {
		long val; 
		int idx;
		
		Tuple() {val = idx = 0;}
		
		Tuple(long a, int b) {
			this.val = a;
			this.idx = b;
		}
		
		@Override
		public int compareTo(CFPSOptimized.Tuple o) {
			if (this.val > o.val)
				return 1;
			if (this.val < o.val)
				return -1;
			return 0;
		}
		
	}
	
	static class tupIdx2Comparator implements Comparator<Tuple> {
		@Override
		public int compare(CFPSOptimized.Tuple o1, CFPSOptimized.Tuple o2) {
			if (o1.idx > o2.idx)
				return 1;
			if (o1.idx < o2.idx)
				return -1;
			return 0;
		}
	}
	
	// Code from geeksforgeeks.org
	// Time Complexity: O(sqrt(n))
	static boolean isPrime(int n) {
		// Corner cases
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;

		// This is checked so that we can skip
		// middle five numbers in below loop
		if (n % 2 == 0 || n % 3 == 0)
			return false;

		for (int i = 5; i * i <= n; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;

		return true;
	}

	static int hash(String s, int arrayLen) {
		return (s.hashCode() & 0x7fffffff) % arrayLen;
	}

	static class FastReader {
		// The input stream is read and stored into a buffer which can be used to
		// retrieve
		// data whenever required. This reduces the # of IO operations thus ensuring
		// fast input.

		// BufferedReader for input buffering and StringTokenizer for splitting the
		// input
		// line by line into string tokens which can be parsed manually for the desired
		// data.
		private BufferedReader bfr;
		private StringTokenizer st;

		public FastReader() {
			// Passing an InputStreamReader to the BufferedReader constructor (using the
			// console
			// input stream).
			bfr = new BufferedReader(new InputStreamReader(System.in));
		}

		// Method to return the next String token which can be parsed further.
		String next() {
			// Initializing (if required)/Resetting by creating tokens from the
			// next line.
			// If there are no more elements from the current line:
			if (st == null || !st.hasMoreElements()) {
				try {
					// Creating tokenizer using the next line.
					st = new StringTokenizer(bfr.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		// Parsing the next token to an Integer.
		int nextInt() {
			return Integer.parseInt(next());
		}

		// Parsing the next token to a Long.
		long nextLong() {
			return Long.parseLong(next());
		}

		// Parsing the next token to a Double.
		double nextDouble() {
			return Double.parseDouble(next());
		}

		// Parsing the next token to a Character.
		char nextChar() {
			return next().toCharArray()[0];
		}

		// Reading a whole string.
		String nextString() {
			return next();
		}

		// Parsing the input tokens to an integer array of
		// the specified size.
		int[] nextIntArray(int n) {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++)
				arr[i] = nextInt();
			return arr;
		}

		// Parsing the input token to a double array of
		// the specified size.
		double[] nextDoubleArray(int n) {
			double[] arr = new double[n];
			for (int i = 0; i < arr.length; i++)
				arr[i] = nextDouble();
			return arr;
		}

		long[] nextLongArray(int n) {
			long[] arr = new long[n];
			for (int i = 0; i < n; i++)
				arr[i] = nextLong();
			return arr;
		}

		public char[] nextCharArray(int n) {
			char[] chars = new char[n];
			for (int i = 0; i < n; i++)
				chars[i] = fr.nextChar();
			return chars;
		}
	}
}