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:
//package codeforces.deltix;

import java.io.*;
import java.util.*;

public class C {
    static InputReader in;
    static PrintWriter out;

    public static void main(String[] args) {
        //initReaderPrinter(true);
        initReaderPrinter(false);
        solve(in.nextInt());
        //solve(1);        
    }

    static class Node {
        int v, id;
        List<Node> child;

        Node(int v, int id) {
            this.v = v;
            this.id = id;
            child = new ArrayList<>();
        }
    }

    static String[] ans;
    static void solve(int testCnt) {
        for (int testNumber = 0; testNumber < testCnt; testNumber++) {
            int n = in.nextInt();
            int[] a = in.nextIntArrayPrimitive(n);
            ans = new String[n];
//            Node root = new Node(a[0], 0);
//            for(int i = 1; i < n; i++) {
//                insert(root, a[i], i);
//            }
//            dfs(root, new StringBuilder());
//            for(int i = 0; i < n; i++) {
//                out.println(ans[i]);
//            }
            ArrayDeque<Integer> q = new ArrayDeque<>();
            q.addLast(a[0]);
            for(int i = 1; i < n; i++) {
                printBuffer(q);
                if(a[i] == 1) {
                    q.addLast(a[i]);
                }
                else {
                    while(q.size() > 0 && q.peekLast() + 1 != a[i]) {
                        q.removeLast();
                    }
                    if(q.size() > 0) {
                        q.removeLast();
                    }
                    q.add(a[i]);
                }
            }
            printBuffer(q);
        }
        out.close();
    }

    static void printBuffer(ArrayDeque<Integer> q) {
        int sz = q.size();
        StringBuilder buffer = new StringBuilder();
        for(int j = 0; j < sz; j++) {
            int v = q.removeFirst();
            buffer.append(v);
            buffer.append(".");
            q.addLast(v);
        }
        buffer.deleteCharAt(buffer.length() - 1);
        out.println(buffer);
    }

    static void insert(Node root, int v, int idx) {
        ArrayDeque<Node> q = new ArrayDeque<>();
        q.addLast(root);
        while(q.size() > 0) {
            Node curr = q.removeFirst();
            if(curr.child.size() + 1 == v) {
                curr.child.add(new Node(v, idx));
                break;
            }
            else {
                for(Node next : curr.child) {
                    q.addLast(next);
                }
            }
        }
    }

    static void dfs(Node curr, StringBuilder buffer) {
        if(buffer.length() > 0) {
            buffer.append("." + curr.v);
        }
        else {
            buffer.append(curr.v);
        }
        ans[curr.id] = buffer.toString();
        for(Node next : curr.child) {
            dfs(next, new StringBuilder(buffer));
        }
    }

    static void initReaderPrinter(boolean test) {
        if (test) {
            try {
                in = new InputReader(new FileInputStream("src/input.in"));
                out = new PrintWriter(new FileOutputStream("src/output.out"));
            } catch (IOException e) {
                e.printStackTrace();
            }
        } else {
            in = new InputReader(System.in);
            out = new PrintWriter(System.out);
        }
    }

    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        InputReader(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream), 32768);
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

        String next() {
            while (st == null || !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());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        Integer[] nextIntArray(int n) {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int[] nextIntArrayPrimitive(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int[] nextIntArrayPrimitiveOneIndexed(int n) {
            int[] a = new int[n + 1];
            for (int i = 1; i <= n; i++) a[i] = nextInt();
            return a;
        }

        Long[] nextLongArray(int n) {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        long[] nextLongArrayPrimitive(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        long[] nextLongArrayPrimitiveOneIndexed(int n) {
            long[] a = new long[n + 1];
            for (int i = 1; i <= n; i++) a[i] = nextLong();
            return a;
        }

        String[] nextStringArray(int n) {
            String[] g = new String[n];
            for (int i = 0; i < n; i++) g[i] = next();
            return g;
        }
    }
}