Problem: Phoenix loves beautiful arrays. An array is beautiful if all its subarrays of length k have the same sum. A subarray of an array is any sequence of consecutive elements.

Phoenix currently has an array a of length n. He wants to insert some number of integers, possibly zero, into his array such that it becomes beautiful. The inserted integers must be between 1 and n inclusive. Integers may be inserted anywhere (even before the first or after the last element), and he is not trying to minimize the number of inserted integers.

Input

The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 50) — the number of test cases.

The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 100).

The second line of each test case contains n space-separated integers (1 ≤ a_i ≤ n) — the array that Phoenix currently has. This array may or may not be already beautiful.

Output

For each test case, if it is impossible to create a beautiful array, print -1. Otherwise, print two lines.

The first line should contain the length of the beautiful array m (n ≤ m ≤ 10^4). You don't need to minimize m.

The second line should contain m space-separated integers (1 ≤ b_i ≤ n) — a beautiful array that Phoenix can obtain after inserting some, possibly zero, integers into his array a. You may print integers that weren't originally in array a.

If there are multiple solutions, print any. It's guaranteed that if we can make array a beautiful, we can always make it with resulting length no more than 10^4.

Example

Input


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


Output


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

Note

In the first test case, we can make array a beautiful by inserting the integer 1 at index 3 (in between the two existing 2s). Now, all subarrays of length k=2 have the same sum 3. There exists many other possible solutions, for example: 

  * 2, 1, 2, 1, 2, 1 
  * 1, 2, 1, 2, 1, 2 



In the second test case, the array is already beautiful: all subarrays of length k=3 have the same sum 5.

In the third test case, it can be shown that we cannot insert numbers to make array a beautiful.

In the fourth test case, the array b shown is beautiful and all subarrays of length k=4 have the same sum 10. There exist other solutions also.

Solution (in JAVA):

Solution:
//package codeforces.round638div2;

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

public class BeautifulArray {
    public static void main(String[] args) {
//        try {
//            FastScanner in = new FastScanner(new FileInputStream("src/input.in"));
//            PrintWriter out = new PrintWriter(new FileOutputStream("src/output.out"));

        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

            int t = in.nextInt();
            solve(t, in, out);
//        } catch (IOException e) {
//            e.printStackTrace();
//        }
    }

    private static void solve(int q, FastScanner in, PrintWriter out) {
        for (int qq = 0; qq < q; qq++) {
            int n = in.nextInt(), k = in.nextInt();
            Integer[] a = in.nextIntArray(n);
            if(n == k) {
                out.println(n);
                for(int i = 0; i < n; i++) {
                    out.print(a[i] + " ");
                }
                out.println();
            }
            else {
                Set<Integer> set = new HashSet<>();
                for(int v : a) {
                    set.add(v);
                }
                if(set.size() > k) {
                    out.println(-1);
                }
                else {
                    List<Integer> list = new ArrayList<>();
                    while(list.size() < k) {
                        for(int v : set) {
                            list.add(v);
                            if(list.size() == k) {
                                break;
                            }
                        }
                    }
                    for(int i = 0; i < n - 1; i++) {
                        for(int j = 0; j < k; j++) {
                            list.add(list.get(j));
                        }
                    }
                    out.println(list.size());
                    for(int i = 0; i < list.size(); i++) {
                        out.print(list.get(i) + " ");
                    }
                    out.println();
                }
            }
        }
        out.close();
    }

    private static class BinarySearch {
        // return the right most index i such that a[i] <= t
        private static int rightMostUpperBound(Integer[] a, int t) {
            int l = 0, r = a.length - 1;
            while (l < r - 1) {
                int mid = l + (r - l) / 2;
                if (a[mid] <= t) {
                    l = mid;
                } else {
                    r = mid - 1;
                }
            }
            if (a[r] <= t) {
                return r;
            } else if (a[l] <= t) {
                return l;
            }
            return -1;
        }

        // return the left most index i such that a[i] >= t
        private static int leftMostLowerBound(Integer[] a, int t) {
            int l = 0, r = a.length - 1;
            while (l < r - 1) {
                int mid = l + (r - l) / 2;
                if (a[mid] >= t) {
                    r = mid;
                } else {
                    l = mid + 1;
                }
            }
            if (a[l] >= t) {
                return l;
            } else if (a[r] >= t) {
                return r;
            }
            return -1;
        }

        // Find the start and end index of a given t in a sorted(ascending) array; return [-1, -1] is t is not found
        private static int[] searchRange(Integer[] a, int t) {
            int first = firstPos(a, t);
            int last = firstPos(a, t + 1) - 1;
            if (first <= last) {
                return new int[]{first, last};
            }
            return new int[]{-1, -1};
        }

        private static int firstPos(Integer[] a, int t) {
            int first = a.length;
            int l = 0, r = a.length - 1;
            while (l <= r) {
                int mid = l + (r - l) / 2;
                if (a[mid] >= t) {
                    first = mid;
                    r = mid - 1;
                } else {
                    l = mid + 1;
                }
            }
            return first;
        }
    }

    private static class NumberTheory {
        private static long modularAdd(long a, long b, int mod) {
            long sum = a + b;
            if (sum >= mod) {
                sum -= mod;
            }
            return sum;
        }

        private static long modularSubtract(long a, long b, int mod) {
            long diff = a - b;
            if (diff < 0) {
                diff += mod;
            }
            return diff;
        }

        private static long fastModPow(long x, int n, long mod) {
            if (n == 0) {
                return 1;
            }
            long coeff = 1, base = x;
            while (n > 1) {
                if (n % 2 != 0) {
                    coeff = (coeff * base % mod);
                }
                base = (base * base % mod);
                n = n / 2;
            }
            long res = coeff * base % mod;
            return res;
        }

        private static long gcd(long a, long b) {
            if (b == 0) {
                return a;
            }
            return gcd(b, a % b);
        }

        private static long[] factorial(int n, long mod) {
            long[] factor = new long[n + 1];
            factor[0] = 1;
            for (int i = 1; i <= n; i++) {
                factor[i] = factor[i - 1] * i % mod;
            }
            return factor;
        }

        private static long modInv_EEA(long a, long mod) {
            long[] res = solveGcdEquation(a, mod);
            if (res[2] != 1) {
                // a and m are not coprime, modular inverse of a by m does not exist.
                return -1;
            }
            return (res[0] % mod + mod) % mod;
        }

        private static long[] solveGcdEquation(long a, long b) {
            if (b == 0) {
                return new long[]{1, 0, a};
            }
            long[] res = solveGcdEquation(b, a % b);
            return new long[]{res[1], res[0] - (a / b) * res[1], res[2]};
        }

        private static long binomialCoefficient(long[] factorial, long n, long k, long mod) {
            return factorial[(int) n] * modInv_EEA(factorial[(int) k], mod) % mod
                    * modInv_EEA(factorial[(int) (n - k)], mod) % mod;
        }
    }

    private static class Graph {
        private static int UNVISITED = 0;
        private static int VISITING = -1;
        private static int VISITED = 1;

        private static Stack<Integer> topologicalSort(Set<Integer>[] g) {
            Stack<Integer> stack = new Stack<>();
            int[] state = new int[g.length];

            for (int node = 0; node < g.length; node++) {
                if (!topoSortHelper(g, stack, state, node)) {
                    return null;
                }
            }
            return stack;
        }

        private static boolean topoSortHelper(Set<Integer>[] g, Stack<Integer> stack, int[] state, int currNode) {
            if (state[currNode] == VISITED) {
                return true;
            } else if (state[currNode] == VISITING) {
                return false;
            }
            state[currNode] = VISITING;
            for (int neighbor : g[currNode]) {
                if (!topoSortHelper(g, stack, state, neighbor)) {
                    return false;
                }
            }
            state[currNode] = VISITED;
            stack.push(currNode);
            return true;
        }
    }

    private static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream));
            } 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;
        }

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