Problem: Vasya has written some permutation p_1, p_2, …, p_n of integers from 1 to n, so for all 1 ≤ i ≤ n it is true that 1 ≤ p_i ≤ n and all p_1, p_2, …, p_n are different. After that he wrote n numbers next_1, next_2, …, next_n. The number next_i is equal to the minimal index i < j ≤ n, such that p_j > p_i. If there is no such j let's let's define as next_i = n + 1.

In the evening Vasya went home from school and due to rain, his notebook got wet. Now it is impossible to read some written numbers. Permutation and some values next_i are completely lost! If for some i the value next_i is lost, let's say that next_i = -1.

You are given numbers next_1, next_2, …, next_n (maybe some of them are equal to -1). Help Vasya to find such permutation p_1, p_2, …, p_n of integers from 1 to n, that he can write it to the notebook and all numbers next_i, which are not equal to -1, will be correct. 

Input

The first line contains one integer t — the number of test cases (1 ≤ t ≤ 100 000).

Next 2 ⋅ t lines contains the description of test cases,two lines for each. The first line contains one integer n — the length of the permutation, written by Vasya (1 ≤ n ≤ 500 000). The second line contains n integers next_1, next_2, …, next_n, separated by spaces (next_i = -1 or i < next_i ≤ n + 1).

It is guaranteed, that the sum of n in all test cases doesn't exceed 500 000.

In hacks you can only use one test case, so T = 1.

Output

Print T lines, in i-th of them answer to the i-th test case.

If there is no such permutations p_1, p_2, …, p_n of integers from 1 to n, that Vasya could write, print the only number -1.

In the other case print n different integers p_1, p_2, …, p_n, separated by spaces (1 ≤ p_i ≤ n). All defined values of next_i which are not equal to -1 should be computed correctly p_1, p_2, …, p_n using defenition given in the statement of the problem. If there exists more than one solution you can find any of them.

Example

Input


6
3
2 3 4
2
3 3
3
-1 -1 -1
3
3 4 -1
1
2
4
4 -1 4 5


Output


1 2 3
2 1
2 1 3
-1
1
3 2 1 4

Note

In the first test case for permutation p = [1, 2, 3] Vasya should write next = [2, 3, 4], because each number in permutation is less than next. It's easy to see, that it is the only satisfying permutation.

In the third test case, any permutation can be the answer because all numbers next_i are lost.

In the fourth test case, there is no satisfying permutation, so the answer is -1.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class DFSBased {

	public static class Reader {
		
		static BufferedReader reader;
		static StringTokenizer tokenizer;
		
		static void init(InputStream input) {
			reader = new BufferedReader(new InputStreamReader(input));
			tokenizer = new StringTokenizer("");
		}
		
		static String next() throws IOException {
			while(!tokenizer.hasMoreTokens()) {
				tokenizer = new StringTokenizer(reader.readLine());
			}
			return tokenizer.nextToken();
		}
		
		static int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
		
		static long nextLong() throws IOException {
			return Long.parseLong(next());
		}
		
	}
		
	private class Node implements Comparable<Node>{
		public final int ix;
		public int nexti;
		public int parents;
		public int val;
		public boolean visited;
		public ArrayList<Node> children;
		
		public Node(int i, int ni) {
			// TODO Auto-generated constructor stub
			ix=i;
			nexti=ni;
			val=0;
			parents=0;
			visited = false;
			children = new ArrayList<>();
		}
		
		public void addChild(Node nd) {
			children.add(nd);
			nd.parents++;
		}

		public void addChild(int i) {
			children.add(nodes[i]);
		}
		
		public void dfs() {
			visited=true;
			int l = children.size();
			for(int i=0; i<l; i++) {
				Node child = children.get(i);
				if(!child.visited) {
					child.dfs();
				}
			}
			solution.add(this);
		}
		
		@Override
		public String toString() {
			// TODO Auto-generated method stub
			return val+"";
		}
		
		@Override
		public int compareTo(Node o) {
			// TODO Auto-generated method stub
			return 0;
		}
	}
	
	public final int n;
	public long ans;
	public static int[] next;
	public final Node[] nodes;
	public final ArrayList<Node> solution;
	
	public DFSBased(int size) {
		
		n=size;
		nodes = new Node[n];
		for(int i=0; i<n; i++) {
			nodes[i] = new Node(i, next[i]);
			
		}
		solution = new ArrayList<>();
	}
	
	public static void main(String[] args) throws IOException {
		
		Reader.init(System.in);
		
		PrintWriter out = new PrintWriter(System.out);
		
		int tc=1;
		tc = Reader.nextInt();
		
		main_loop:
		for(int tn=0; tn<tc; tn++) {
			
			int n = Reader.nextInt();
			
			next = new int[n];
			
			for(int i=0; i<n; i++) {
				next[i] = Reader.nextInt()-1;
			}

			DFSBased obj = new DFSBased(n);
			
			if(!obj.buildTree()) {
				out.println(-1);
				continue;
			}
			
			obj.solve1();

			for(int i=0; i<n; i++) {
				obj.solution.get(i).val = i+1;
			}
			
			for(int i=0; i<n; i++) {
				out.print(obj.nodes[i].val+" ");
			}
			
			out.println();
						
		}
		
		out.flush();
		out.close();
	}

	private void solve1() {
		// TODO Auto-generated method stub
		PriorityQueue<Node> pq = new PriorityQueue<Node>();
		
		for(int i=0; i<n; i++) {
			Node cur = nodes[i];
			if(cur.parents==0) {
				pq.add(cur);
			}
		}
		
		while(!pq.isEmpty()) {
			Node cur = pq.poll();
			solution.add(cur);
			int ch = cur.children.size();
			for(int i=0; i<ch; i++) {
				Node child = cur.children.get(i);
				child.parents--;
				if(child.parents==0) {
					pq.add(child);
				}
			}
		}
		
		Collections.reverse(solution);
		
		

	}

	private boolean buildTree() {
		Node[] stack = new Node[n];
		int l=0;
		
		for(int i=0; i<n; i++) {
			while(l>0 && stack[l-1].nexti==i) {
				l--;
			}
			int limit = n;
			if(l>0) {
				limit = stack[l-1].nexti;
			}
			if(nodes[i].nexti==-2) {
				nodes[i].nexti=limit;
			}
			if(nodes[i].nexti>limit) {
				return false;
			}
			if(l==0) {
				if(nodes[i].nexti!=n) {					
					nodes[nodes[i].nexti].addChild(nodes[i]);
				}
			}
			else {
				if(limit==nodes[i].nexti) {
					stack[l-1].addChild(nodes[i]);
				}
				else {
					nodes[nodes[i].nexti].addChild(nodes[i]);
				}
			}
			stack[l] = nodes[i];
			l++;
		}
		
		return true;
	}

	private void solve() {
		for(int i=0; i<n; i++) {
			if(!nodes[i].visited) {
				nodes[i].dfs();
			}
		}
//		Collections.reverse(solution);
		
		int l = solution.size();
		
		if(l!=n) {
			System.exit(-1);
		}
		
		
	}

}