Problem: <image>

William is a huge fan of planning ahead. That is why he starts his morning routine by creating a nested list of upcoming errands.

A valid nested list is any list which can be created from a list with one item "1" by applying some operations. Each operation inserts a new item into the list, on a new line, just after one of existing items a_1  .  a_2  .  a_3  .   ⋅⋅⋅   .\,a_k and can be one of two types: 

  1. Add an item a_1  .  a_2  .  a_3  .  ⋅⋅⋅  .  a_k  .  1 (starting a list of a deeper level), or 
  2. Add an item a_1  .  a_2  .  a_3  .  ⋅⋅⋅  .  (a_k + 1) (continuing the current level). 

Operation can only be applied if the list does not contain two identical items afterwards. And also, if we consider every item as a sequence of numbers, then the sequence of items should always remain increasing in lexicographical order. Examples of valid and invalid lists that are shown in the picture can found in the "Notes" section.

When William decided to save a Word document with the list of his errands he accidentally hit a completely different keyboard shortcut from the "Ctrl-S" he wanted to hit. It's not known exactly what shortcut he pressed but after triggering it all items in the list were replaced by a single number: the last number originally written in the item number.

William wants you to help him restore a fitting original nested list.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3), which is the number of lines in the list.

Each of the next n lines contains a single integer a_i (1 ≤ a_i ≤ n), which is what remains of William's nested list.

It is guaranteed that in each test case at least one fitting list exists.

It is guaranteed that the sum of values n across all test cases does not exceed 10^3.

Output

For each test case output n lines which represent a valid nested list, which could become the data provided to you by William.

If there are multiple answers, print any.

Example

Input


2
4
1
1
2
3
9
1
1
1
2
2
1
2
1
2


Output


1
1.1
1.2
1.3
1
1.1
1.1.1
1.1.2
1.2
1.2.1
2
2.1
2.2

Note

In the second example test case one example of a fitting list is:

1

1.1

1.1.1

1.1.2

1.2

1.2.1

2

2.1

2.2

This list can be produced by using the sequence of operations shown below: <image>

  1. Original list with a single item 1. 
  2. Insert item 2 by using the insertion operation of the second type after item 1. 
  3. Insert item 1.1 by using the insertion operation of the first type after item 1. 
  4. Insert item 1.2 by using the insertion operation of the second type after item 1.1. 
  5. Insert item 1.1.1 by using the insertion operation of the first type after item 1.1. 
  6. Insert item 1.1.2 by using the insertion operation of the second type after item 1.1.1. 
  7. Insert item 1.2.1 by using the insertion operation of the first type after item 1.2. 
  8. Insert item 2.1 by using the insertion operation of the first type after item 2. 
  9. Insert item 2.2 by using the insertion operation of the second type after item 2.1. 

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import static java.lang.Math.PI;
import static java.lang.System.in;
import static java.lang.System.out;
import static java.lang.System.err;
/*
11
1
1
1
2
2
1
2
1
2
2
3
*/
public class C {    
    static public void main(String... args) throws Exception {   
        Foster sc = new Foster();
        PrintWriter p = new PrintWriter(out);
        int t = sc.nextInt();
        while(t--!=0){
            int n = sc.nextInt();
            int a[] = sc.intArray(n);
            ArrayList<ArrayList<Integer>> arr = new ArrayList<>();
            for(int i = 0; i < n; i++){
                ArrayList<Integer> temp = new ArrayList<>();
                if(i-1 < 0){
                    temp.add(1);
                }
                else{
                    ArrayList<Integer> inner = arr.get(i-1);
                    int last = inner.get(inner.size()-1);
                    ArrayDeque<Integer> q = new ArrayDeque<>();
                    for(int j : inner){
                        q.addLast(j);
                    }
                    // if current is one greater then increment
                    if(last+1 == a[i]){
                        q.pollLast();
                        q.addLast(a[i]);
                    }
                    // move the chain
                    else if(a[i]==1){
                        q.addLast(a[i]);
                    }
                    // move back
                    else{
                        while(!q.isEmpty() && a[i]-q.peekLast() != 1){
                            q.pollLast();
                        }
                        if(q.isEmpty()) q.addLast(a[i]);
                        else{
                            q.pollLast();
                            q.addLast(a[i]);
                        }
                    }
                    // transfer queue to list
                    while(!q.isEmpty()){
                        temp.add(q.pollFirst());
                    }
                }
                arr.add(temp);
            }

            // output answer
            for(int i = 0; i < arr.size(); i++){
                p.print(arr.get(i).get(0));
                for(int j = 1; j < arr.get(i).size(); j++){
                    p.print("." + arr.get(i).get(j));
                }
                p.println();
            }
        }
        p.close();
    }
    
    static long[] sort(long a[]){
        ArrayList<Long> arr = new ArrayList<>();
        for(long i : a){
            arr.add(i);
        }
        Collections.sort(arr);
        for(int i = 0; i < arr.size(); i++){
            a[i] = arr.get(i);
        }
        return a;
    }
    static int[] sort(int a[]){
        ArrayList<Integer> arr = new ArrayList<>();
        for(int i : a){
            arr.add(i);
        }
        Collections.sort(arr);
        // Collections.reverse(arr);
        for(int i = 0; i < arr.size(); i++){
            a[i] = arr.get(i);
        }
        return a;
    }
    
    
/* 
*/    
/*
1. Check overflow in pow function or in general
2. Check indices of read array function
3. Think of an easier solution because the problems you solve are always easy
4. Check iterator of loop
5. If still doesn't work, then jump from the 729th floor 'coz "beta tumse na ho paayega"

    Move to top!!
*/
    static class Foster {
        BufferedReader br = new BufferedReader(new InputStreamReader(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());
        }
        int[] intArray(int n) {                   // Check indices
            int arr[] = new int[n];
            for(int i = 0; i < n; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }
        long[] longArray(int n) {                 // Check indices
            long arr[] = new long[n];
            for(int i = 0; i < n; i++) {
                arr[i] = nextLong();
            }
            return arr;
        }
        int[] getBits(int n) {                   //in Reverse Order
            int a[] = new int[31];
            for(int i = 0; i < 31; i++) {
                if(((1<<i) & n) != 0)
                    a[i] = 1;
            }
            return a;
        }
        static long pow(long... a) {
            long mod = Long.MAX_VALUE;
            if(a.length == 3)   mod = a[2];
            long res = 1;
            while(a[1] > 0) {
                if((a[1] & 1) == 1)
                    res = (res * a[0]) % mod;
                a[1] /= 2;
                a[0] = (a[0] * a[0]) % mod;
            }
            return res;
        }
    }
}