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 Q5 {
	static int INF = (int)(1e9);
	static long mod = (long)(1e9)+7;
	static long mod2 = 998244353;
	static long[] segtree;
	static char[] curans;
	static long ans;	
	static String S;
	static ArrayList<Integer>[] graph;
	static boolean[] vis;
	static int[] a;
	static int N;
	static long K;
	static long[] fact;
	static ArrayList<Long> pos;
	static ArrayList<Long> neg;
	static long[] max;
	static int[] dp;
	public static void main(String[] args) {  //Cash out
		/**/
		FastScanner I = new FastScanner(); //Input
		OutPut O = new OutPut(); //Output
		int T = I.nextInt();
		StringBuilder ret = new StringBuilder();
		while (T-- > 0) {
			int N = I.nextInt();
			long[] a = new long[N + 1];
			PriorityQueue<Long> sort = new PriorityQueue<Long>(Collections.reverseOrder());
			long sum = 0;
			int ans = 1;
			for (int i = 1; i <= N; i++) {
				a[i] = I.nextLong();
				sort.add(a[i]);
				sum += a[i];
			}
			
			long curmax = sort.poll();
			sum -= curmax;
			while (!sort.isEmpty()) {
				if (sum >= curmax) {
					ans++;
					curmax = sort.poll();
					sum -= curmax;
				}else {
					break;
				}
			}
			ret.append(ans + "\n");
			for (int i = 1; i <= N; i++) {
				if (a[i] >= curmax) {
					ret.append(i + " ");
				}
			} 
			ret.append("\n");
		}
		O.pln(ret);
	}
	public static double[][] matrix_exp(double[][] a, long exp, long mod) {
		int R = a.length;
		int C = a[0].length;
		double[][] ans = new double[R][C];
		
		boolean mult_yet = false;
		
		while (exp > 0) {
			if (exp % 2 == 1) {
				if (!mult_yet) {
					mult_yet = true;
					for (int i = 0; i < R; i++) {
						for (int j = 0; j < C; j++) {
							ans[i][j] = a[i][j]; 
						}
					}
				}else {
					double[][] new_ans = mult(ans, a, mod);
					for (int i = 0; i < R; i++) {
						for (int j = 0; j < C; j++) {
							ans[i][j] = new_ans[i][j]; 
						}
					}
				}
			}
			
			double[][] new_a = mult(a, a, mod); // a = a^2 (binary exponentiation on matrices)
			for (int i = 0; i < R; i++) {
				for (int j = 0; j < C; j++) {
					a[i][j] = new_a[i][j]; 
				}
			}
			exp /= 2;
		}
		
		return ans;
	}
	public static double[][] mult(double[][] a, double[][] b, long mod) {
		int r1 = a.length;
		int c1 = a[0].length;
		int r2 = b.length;
		int c2 = b[0].length;
		//Requires: c1 = r2
		
		double[][] ans = new double[r1][c2];
		for (int r = 0; r < r1; r++) {
			for (int c = 0; c < c2; c++) {
				//Dot product of (a[r])^T and b[c] (as vectors in R^n (n = c1 = r2))
				double[] col_vector = new double[r2];
				double[] row_vector = new double[c1];
				
				for (int i = 0; i < r2; i++) {
					col_vector[i] = b[i][c];
				}
				for (int i = 0; i < c1; i++) {
					row_vector[i] = a[r][i]; 
				}
				
				ans[r][c] = dot_product(row_vector, col_vector, mod);
			}
		}
		return ans;
	}
	
	public static double dot_product(double[] a, double[] b, long mod) {
		double ans = 0;
		int N = a.length; //Requires: a and b are both vectors in R^n
		for (int i = 0; i < N; i++) {
			ans += a[i] * b[i];
		}
		return ans;
	}
	public static double max(double a, double b) {return Math.max(a, b);}
	public static double min(double a, double b) {return Math.min(a, b);}
	public static long min(long a, long b) {return Math.min(a,b);}
	public static long max(long a, long b) {return Math.max(a,b);}
	public static int min(int a, int b) {return Math.min(a,b);}
	public static int max(int a, int b) {return Math.max(a,b);}
	public static long abs(long x) {return Math.abs(x);}
	public static long abs(int x) {return Math.abs(x);}
	public static long ceil(long num, long den) {long ans = num/den; if (num%den!=0) 
	ans++; return ans;}
	public static long GCD(long a, long b) {
		if (a==0||b==0) return max(a,b);
		return GCD(min(a,b),max(a,b)%min(a,b));
	}
	public static long FastExp(long base, long exp, long mod) {
		long ans=1;
		while (exp>0) {
			if (exp%2==1) ans*=base;
			exp/=2;
			base*=base;
			base%=mod;
			ans%=mod;
		}
		return ans;
	}
	public static long ModInv(long num,long mod) {return FastExp(num,mod-2,mod);}
	public static int pop(long x) { //Returns number of bits within a number
		int cnt = 0;
		while (x>0) {
			if (x%2==1) cnt++;
			x/=2;
		}
		return cnt;
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());};
		double nextDouble() {return Double.parseDouble(next());}
	}
	static class OutPut{
		PrintWriter w = new PrintWriter(System.out);
		void pln(double x) {w.println(x);w.flush();}
		void pln(boolean x) {w.println(x);w.flush();}
		void pln(int x) {w.println(x);w.flush();}
		void pln(long x) {w.println(x);w.flush();}
		void pln(String x) {w.println(x);w.flush();}
		void pln(char x) {w.println(x);w.flush();}
		void pln(StringBuilder x) {w.println(x);w.flush();}
		void pln(BigInteger x) {w.println(x);w.flush();}
		void p(int x) {w.print(x);w.flush();}
		void p(long x) {w.print(x);w.flush();}
		void p(String x) {w.print(x);w.flush();}
		void p(char x) {w.print(x);w.flush();}
		void p(StringBuilder x) {w.print(x);w.flush();}
		void p(BigInteger x) {w.print(x);w.flush();}
		void p(double x) {w.print(x);w.flush();}
		void p(boolean x) {w.print(x);w.flush();}
	}
}