Problem: A permutation is a sequence of n integers from 1 to n, in which all numbers occur exactly once. For example, [1], [3, 5, 2, 1, 4], [1, 3, 2] are permutations, and [2, 3, 2], [4, 3, 1], [0] are not.

Polycarp was presented with a permutation p of numbers from 1 to n. However, when Polycarp came home, he noticed that in his pocket, the permutation p had turned into an array q according to the following rule: 

  * q_i = max(p_1, p_2, …, p_i). 



Now Polycarp wondered what lexicographically minimal and lexicographically maximal permutations could be presented to him.

An array a of length n is lexicographically smaller than an array b of length n if there is an index i (1 ≤ i ≤ n) such that the first i-1 elements of arrays a and b are the same, and the i-th element of the array a is less than the i-th element of the array b. For example, the array a=[1, 3, 2, 3] is lexicographically smaller than the array b=[1, 3, 4, 2].

For example, if n=7 and p=[3, 2, 4, 1, 7, 5, 6], then q=[3, 3, 4, 4, 7, 7, 7] and the following permutations could have been as p initially: 

  * [3, 1, 4, 2, 7, 5, 6] (lexicographically minimal permutation); 
  * [3, 1, 4, 2, 7, 6, 5]; 
  * [3, 2, 4, 1, 7, 5, 6]; 
  * [3, 2, 4, 1, 7, 6, 5] (lexicographically maximum permutation). 



For a given array q, find the lexicographically minimal and lexicographically maximal permutations that could have been originally presented to Polycarp.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).

The second line of each test case contains n integers q_1, q_2, …, q_n (1 ≤ q_i ≤ n).

It is guaranteed that the array q was obtained by applying the rule from the statement to some permutation p.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output two lines: 

  * on the first line output n integers — lexicographically minimal permutation that could have been originally presented to Polycarp; 
  * on the second line print n integers — lexicographically maximal permutation that could have been originally presented to Polycarp; 

Example

Input


4
7
3 3 4 4 7 7 7
4
1 2 3 4
7
3 4 5 5 5 7 7
1
1


Output


3 1 4 2 7 5 6 
3 2 4 1 7 6 5 
1 2 3 4 
1 2 3 4 
3 4 5 1 2 7 6 
3 4 5 2 1 7 6 
1 
1 

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class E {

	public static Reader sc = new Reader();
	//public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

	public static void main(String[] args) throws IOException {
		//BufferedReader br = new BufferedReader(new FileReader("input.in"));

		int t = sc.nextInt();
		while (t-- > 0) {
			TreeSet<Integer> tsMin = new TreeSet<Integer>();
			TreeSet<Integer> tsMax = new TreeSet<Integer>();

			int n = sc.nextInt();
			for (int i = 1; i <=n; i++) {
				 tsMin.add(i);
				 tsMax.add(i);
			}
			int cm = -1;
			
			int[] min = new int[n];
			int[] max = new int[n];
			int[] a = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = sc.nextInt();
				if (a[i] > cm) {
					max[i] = a[i];
					min[i] = a[i];
					tsMin.remove(a[i]);
					tsMax.remove(a[i]);
					cm = a[i];
				} 
			}
			for (int i = 0; i < n; i++) {
				if (min[i] == 0) {
					min[i] = tsMin.pollFirst();
				}
			}
			cm = -1;
			for (int i = 0; i < n; i++) {
				if (a[i] > cm) {
					cm = a[i];
				}
				else if (max[i] == 0) {
					int cur = tsMax.floor(cm-1);
					max[i] = cur;
					tsMax.remove(cur);
				}
			}
			
			
			for (int x: min) out.print(x + " ");
			out.println();
			for (int x: max) out.print(x + " ");
			out.println();
		}

		out.close();
	}

	static void mergeSort(long[] a) {
		if (a.length < 2)
			return;
		long[] x = new long[a.length / 2];
		long[] y = new long[a.length - a.length / 2];
		for (int i = 0; i < x.length; i++)
			x[i] = a[i];
		for (int i = 0; i < y.length; i++)
			y[i] = a[i + x.length];
		mergeSort(x);
		mergeSort(y);
		int i = 0, j = 0, k = 0;
		while (j < x.length && k < y.length) {
			if (x[j] < y[k]) {
				a[i] = x[j];
				j++;
			} else {
				a[i] = y[k];
				k++;
			}
			i++;
		}
		while (j < x.length) {
			a[i] = x[j];
			j++;
			i++;
		}
		while (k < y.length) {
			a[i] = y[k];
			k++;
			i++;
		}
	}

	static long ceil(long a, long b) {
		return (a + b - 1) / b;
	}

	static long gcd(long a, long b) {
		while (b != 0) {
			long t = a;
			a = b;
			b = t % b;
		}
		return a;
	}

	static long powMod(long base, long exp, long mod) {
		if (base == 0 || base == 1)
			return base;
		if (exp == 0)
			return 1;
		if (exp == 1)
			return base % mod;
		long R = powMod(base, exp / 2, mod) % mod;
		R *= R;
		R %= mod;
		if ((exp & 1) == 1) {
			return base * R % mod;
		} else
			return R % mod;
	}

	static long pow(long base, long exp) {
		if (base == 0 || base == 1)
			return base;
		if (exp == 0)
			return 1;
		if (exp == 1)
			return base;
		long R = pow(base, exp / 2);
		if ((exp & 1) == 1) {
			return R * R * base;
		} else
			return R * R;
	}

	static class Reader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;

		public Reader() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public Reader(String file_name) throws IOException {
			din = new DataInputStream(new FileInputStream(file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public String readLine() throws IOException {
			byte[] buf = new byte[64]; // line length 
			int cnt = 0, c;
			while ((c = read()) != -1) {
				if (c == '\n') {
					break;
				}
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, Integer.max(cnt - 1, 0));
		}

		public int nextInt() throws IOException {
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}

		public long nextLong() throws IOException {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}

		public double nextDouble() throws IOException {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (c == '.') {
				while ((c = read()) >= '0' && c <= '9') {
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}

		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}

		private byte read() throws IOException {
			if (bufferPointer == bytesRead)
				fillBuffer();
			return buffer[bufferPointer++];
		}

		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}
	}

}