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:
// package contest;

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.Iterator;
import java.util.TreeMap;

public class E {
	private static PrintWriter out = new PrintWriter(System.out);

	public static void solve() {
		In in = new In();
		int tests = in.ni();
		while (tests-- > 0) {
			int n = in.ni();
			int[] a = in.nia(n);

			int[][] ai = new int[n][];
			for (int i = 0; i < n; i++) {
				ai[i] = new int[] { a[i], i };
			}
			Arrays.sort(ai, (x, y) -> Integer.compare(x[0], y[0]));
			long[] pre = new long[n];
			pre[0] = ai[0][0];
			for(int i = 1; i < n; i++) pre[i] = pre[i - 1] + ai[i][0];

			int lo = 0;
			int hi = n - 1;
			int cannot = -1;
			while (lo <= hi) {
				int mi = lo + (hi - lo) / 2;
				if (!canWin(mi, ai, pre)) {
					cannot = mi;
					lo = mi + 1;
				} else {
					hi = mi - 1;
				}
			}
			int[] ans = new int[n];
			int p = 0;
			for (int i = cannot + 1; i < n; i++)
				ans[p++] = ai[i][1] + 1;
			ans = Arrays.copyOf(ans, p);
			Arrays.sort(ans);
			out.println(p);
			for (int x : ans)
				out.print(x + " ");
			out.println();
		}
	}

	static boolean canWin(int i, int[][] ai, long[] pre) {
		long sum = pre[i];
		int n = ai.length;
		for(int j = i + 1; j < n; j++) {
			if (ai[j][0] > sum) return false;
			sum += ai[j][0];
		}
		return true;
		
	}

	public static void main(String[] args) throws IOException {
		// long start = System.nanoTime();
		solve();
		// System.out.println("Elapsed: " + (System.nanoTime() - start) / 1000000 +
		// "ns");
		out.flush();
	}

	@SuppressWarnings("unused")
	private static class In {
		final private static int BUFFER_SIZE = 1024;
		private byte[] buf;
		private InputStream is;
		private int buflen;
		private int bufptr;

		public In() {
			is = System.in;
			buf = new byte[BUFFER_SIZE];
			buflen = bufptr = 0;
		}

		public In(String input) {
			is = new ByteArrayInputStream(input.getBytes());
			buf = new byte[BUFFER_SIZE];
			buflen = bufptr = 0;
		}

		public int readByte() {
			if (bufptr >= buflen) {
				try {
					buflen = is.read(buf);
				} catch (IOException ioe) {
					throw new InputMismatchException();
				}
				bufptr = 0;
			}
			if (buflen <= 0)
				return -1;
			return buf[bufptr++];
		}

		public boolean isSpaceChar(int c) {
			return !(c >= 33 && c <= 126);
		}

		public int skip() {
			int b;
			while ((b = readByte()) != -1 && isSpaceChar(b))
				;
			return b;
		}

		/* Next character */
		public char nc() {
			return (char) skip();
		}

		/* Next double */
		public double nd() {
			return Double.parseDouble(ns());
		}

		/* Next string */
		public String ns() {
			final StringBuilder sb = new StringBuilder();
			int b = skip();
			while (!isSpaceChar(b)) {
				sb.appendCodePoint(b);
				b = readByte();
			}
			return sb.toString();
		}

		/* Next integer */
		public int ni() {
			boolean minus = false;
			int num = 0;
			int b;
			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
				;
			if (b == '-') {
				minus = true;
				b = readByte();
			}
			while (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
				b = readByte();
			}
			return minus ? -num : num;
		}

		/* Next long */
		public long nl() {
			boolean minus = false;
			long num = 0;
			int b;
			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
				;
			if (b == '-') {
				minus = true;
				b = readByte();
			}
			while (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
				b = readByte();
			}
			return minus ? -num : num;
		}

		/* Next integer 1D array */
		public int[] nia(int n) {
			final int[] arr = new int[n];
			for (int i = 0; i < n; i++)
				arr[i] = ni();
			return arr;
		}

		/* Next long 1D array */
		public long[] nla(int n) {
			final long[] arr = new long[n];
			for (int i = 0; i < n; i++)
				arr[i] = nl();
			return arr;
		}

		/* Next string 1D array */
		public String[] nsa(int n) {
			final String[] arr = new String[n];
			for (int i = 0; i < n; i++)
				arr[i] = ns();
			return arr;
		}

		/* Next char 1D array */
		public char[] nca(int n) {
			final char[] arr = new char[n];
			for (int i = 0; i < n; i++)
				arr[i] = nc();
			return arr;
		}

		/* Next double 1D array */
		public double[] nda(int n) {
			final double[] arr = new double[n];
			for (int i = 0; i < n; i++)
				arr[i] = nc();
			return arr;
		}

		/* Next integer matrix */
		public int[][] nim(int n, int m) {
			final int[][] arr = new int[n][m];
			for (int i = 0; i < n; i++)
				for (int j = 0; j < m; j++)
					arr[i][j] = ni();
			return arr;
		}

		/* Next long matrix */
		public long[][] nlm(int n, int m) {
			final long[][] arr = new long[n][m];
			for (int i = 0; i < n; i++)
				for (int j = 0; j < m; j++)
					arr[i][j] = nl();
			return arr;
		}

		/* Next string matrix */
		public String[][] nsm(int n, int m) {
			final String[][] arr = new String[n][m];
			for (int i = 0; i < n; i++)
				for (int j = 0; j < m; j++)
					arr[i][j] = ns();
			return arr;
		}

		/* Next char matrix */
		public char[][] ncm(int n, int m) {
			final char[][] arr = new char[n][m];
			for (int i = 0; i < n; i++)
				for (int j = 0; j < m; j++)
					arr[i][j] = nc();
			return arr;
		}

		/* Next double matrix */
		public double[][] ndm(int n, int m) {
			final double[][] arr = new double[n][m];
			for (int i = 0; i < n; i++)
				for (int j = 0; j < m; j++)
					arr[i][j] = nd();
			return arr;
		}

		public static void log(Object... o) {
			System.out.println(Arrays.deepToString(o));
		}
	}

}