description: The city where Mocha lives in is called Zhijiang. There are n+1 villages and 2n-1 directed roads in this city. 
  
  There are two kinds of roads:
  
    * n-1 roads are from village i to village i+1, for all 1≤ i ≤ n-1. 
    * n roads can be described by a sequence a_1,…,a_n. If a_i=0, the i-th of these roads goes from village i to village n+1, otherwise it goes from village n+1 to village i, for all 1≤ i≤ n. 
  
  
  
  Mocha plans to go hiking with Taki this weekend. To avoid the trip being boring, they plan to go through every village exactly once. They can start and finish at any villages. Can you help them to draw up a plan? 
  
  Input
  
  Each test contains multiple test cases. 
  
  The first line contains a single integer t (1 ≤ t ≤ 20) — the number of test cases. Each test case consists of two lines.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 10^4) — indicates that the number of villages is n+1.
  
  The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1). If a_i=0, it means that there is a road from village i to village n+1. If a_i=1, it means that there is a road from village n+1 to village i.
  
  It is guaranteed that the sum of n over all test cases does not exceed 10^4.
  
  Output
  
  For each test case, print a line with n+1 integers, where the i-th number is the i-th village they will go through. If the answer doesn't exist, print -1.
  
  If there are multiple correct answers, you can print any one of them.
  
  Example
  
  Input
  
  
  2
  3
  0 1 0
  3
  1 1 0
  
  
  Output
  
  
  1 4 2 3 
  4 1 2 3 
  
  Note
  
  In the first test case, the city looks like the following graph:
  
  <image>
  
  So all possible answers are (1 → 4 → 2 → 3), (1 → 2 → 3 → 4).
  
  In the second test case, the city looks like the following graph:
  
  <image>
  
  So all possible answers are (4 → 1 → 2 → 3), (1 → 2 → 3 → 4), (3 → 4 → 1 → 2), (2 → 3 → 4 → 1).
solution: import static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.*;
  import java.math.*;
  /*
  -> Give your 100%, that's it!
  -> Feynman's Rule To Solve Any Problem:
      1. Read the problem.
      2. Think About It.
      3. Solve it!
  */
  
  public class Template {
      
  	static int mod = 1000000007;
  
  	public static void main(String[] args){
          FastScanner sc = new FastScanner();
          PrintWriter out = new PrintWriter(System.out);
          int yo = sc.nextInt();
          while (yo-- > 0) {
              int n = sc.nextInt();
              int[] a = sc.readInts(n);
              boolean ok = true;
              int now = 1;
              for(int i = 0; i < n; i++){
                  if(i == n-1 && ok){
                      if(a[i] == 0){
                          out.print(n + " ");
                          out.print((n+1) + " ");
                      }
                      else {
                          out.print((n+1) + " ");
                          out.print(n + " ");
                      }
                      break;
                  }
                  else if(a[i] == 1 && ok){
                      ok = false;
                      out.print((n+1) + " ");
                      out.print((now++) + " ");
                  }
                  else {
                      out.print((now++) + " ");
                  }
              }
              out.println();
          }
          out.close();    
  	}
      
     
      
      
      /*
      Source: hu_tao
           Random stuff to try when stuck:
              -if it's 2C then it's dp
              -for combo/probability problems, expand the given form we're interested in
              -make everything the same then build an answer (constructive, make everything 0 then do something)
              -something appears in parts of 2 --> model as graph
              -assume a greedy then try to show why it works
              -find way to simplify into one variable if multiple exist
              -treat it like fmc (note any passing thoughts/algo that could be used so you can revisit them)
              -find lower and upper bounds on answer
              -figure out what ur trying to find and isolate it
              -see what observations you have and come up with more continuations
              -work backwards (in constructive, go from the goal to the start)
              -turn into prefix/suffix sum argument (often works if problem revolves around adjacent array elements)
              -instead of solving for answer, try solving for complement (ex, find n-(min) instead of max)
              -draw something
              -simulate a process
              -dont implement something unless if ur fairly confident its correct
              -after 3 bad submissions move on to next problem if applicable
              -do something instead of nothing and stay organized
              -write stuff down
           Random stuff to check when wa:
              -if code is way too long/cancer then reassess
              -switched N/M
              -int overflow
              -switched variables
              -wrong MOD
              -hardcoded edge case incorrectly
           Random stuff to check when tle:
              -continue instead of break
              -condition in for/while loop bad
           Random stuff to check when rte:
              -switched N/M
              -long to int/int overflow
              -division by 0
              -edge case for empty list/data structure/N=1
        */
  
  	public static class Pair {
  		int x;
  		int y;
  
  		public Pair(int x, int y) {
  			this.x = x;
  			this.y = y;
  		}
  	}
  
  	public static void sort(int[] arr) {
  		ArrayList<Integer> ls = new ArrayList<Integer>();
  		for (int x : arr)
  			ls.add(x);
  		Collections.sort(ls);
  		for (int i = 0; i < arr.length; i++)
  			arr[i] = ls.get(i);
  	}
  
  	public static long gcd(long a, long b) {
  		if (b == 0)
  			return a;
  		return gcd(b, a % b);
  	}
  
  	static boolean[] sieve(int N) {
  		boolean[] sieve = new boolean[N + 1];
  		for (int i = 2; i <= N; i++)
  			sieve[i] = true;
  
  		for (int i = 2; i <= N; i++) {
  			if (sieve[i]) {
  				for (int j = 2 * i; j <= N; j += i) {
  					sieve[j] = false;
  				}
  			}
  		}
  		return sieve;
  	}
  
  	public static long power(long x, long y, long p) {
  		long res = 1L;
  		x = x % p;
  		while (y > 0) {
  			if ((y & 1) == 1)
  				res = (res * x) % p;
  			y >>= 1;
  			x = (x * x) % p;
  		}
  		return res;
  	}
  
  	public static void print(int[] arr) {
  		//for debugging only
  		for (int x : arr)
  			out.print(x + " ");
  		out.println();
  	}
  
  	static class FastScanner {
  		private int BS = 1 << 16;
  		private char NC = (char) 0;
  		private byte[] buf = new byte[BS];
  		private int bId = 0, size = 0;
  		private char c = NC;
  		private double cnt = 1;
  		private BufferedInputStream in;
  
  		public FastScanner() {
  			in = new BufferedInputStream(System.in, BS);
  		}
  
  		public FastScanner(String s) {
  			try {
  				in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
  			} catch (Exception e) {
  				in = new BufferedInputStream(System.in, BS);
  			}
  		}
  
  		private char getChar() {
  			while (bId == size) {
  				try {
  					size = in.read(buf);
  				} catch (Exception e) {
  					return NC;
  				}
  				if (size == -1)
  					return NC;
  				bId = 0;
  			}
  			return (char) buf[bId++];
  		}
  
  		public int nextInt() {
  			return (int) nextLong();
  		}
  
  		public int[] readInts(int N) {
  			int[] res = new int[N];
  			for (int i = 0; i < N; i++) {
  				res[i] = (int) nextLong();
  			}
  			return res;
  		}
  
  		public long[] readLongs(int N) {
  			long[] res = new long[N];
  			for (int i = 0; i < N; i++) {
  				res[i] = nextLong();
  			}
  			return res;
  		}
  
  		public long nextLong() {
  			cnt = 1;
  			boolean neg = false;
  			if (c == NC)
  				c = getChar();
  			for (; (c < '0' || c > '9'); c = getChar()) {
  				if (c == '-')
  					neg = true;
  			}
  			long res = 0;
  			for (; c >= '0' && c <= '9'; c = getChar()) {
  				res = (res << 3) + (res << 1) + c - '0';
  				cnt *= 10;
  			}
  			return neg ? -res : res;
  		}
  
  		public double nextDouble() {
  			double cur = nextLong();
  			return c != '.' ? cur : cur + nextLong() / cnt;
  		}
  
  		public double[] readDoubles(int N) {
  			double[] res = new double[N];
  			for (int i = 0; i < N; i++) {
  				res[i] = nextDouble();
  			}
  			return res;
  		}
  
  		public String next() {
  			StringBuilder res = new StringBuilder();
  			while (c <= 32)
  				c = getChar();
  			while (c > 32) {
  				res.append(c);
  				c = getChar();
  			}
  			return res.toString();
  		}
  
  		public String nextLine() {
  			StringBuilder res = new StringBuilder();
  			while (c <= 32)
  				c = getChar();
  			while (c != '\n') {
  				res.append(c);
  				c = getChar();
  			}
  			return res.toString();
  		}
  
  		public boolean hasNext() {
  			if (c > 32)
  				return true;
  			while (true) {
  				c = getChar();
  				if (c == NC)
  					return false;
  				else if (c > 32)
  					return true;
  			}
  		}
  	}
  
  	//	For Input.txt and Output.txt	
  	//	FileInputStream in = new FileInputStream("input.txt");
  	//	FileOutputStream out = new FileOutputStream("output.txt");
  	//	PrintWriter pw = new PrintWriter(out);
  	//	Scanner sc = new Scanner(in);
  }