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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool isvowel(char c) {
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' ||
      c == 'E' || c == 'I' || c == 'O' || c == 'U')
    return true;
  return false;
}
long long fast_exp(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans *= x;
    n = n >> 1;
    x = x * x;
  }
  return ans;
}
long long mod_exp(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long sod(long long n) {
  long long sum = 0;
  while (n != 0) {
    sum = sum + (n % 10);
    n = n / 10;
  }
  return sum;
}
long long modInverse(long long a, long long m) { return mod_exp(a, m - 2, m); }
long long nCrModpDP(long long n, long long r, long long p) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, r); j > 0; j--) C[j] = (C[j] + C[j - 1]) % p;
  }
  return C[r];
}
long long nCrModpLucas(long long n, long long r, long long p) {
  if (r == 0) return 1;
  long long ni = n % p, ri = r % p;
  return (nCrModpLucas(n / p, r / p, p) * nCrModpDP(ni, ri, p)) % p;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  set<long long> st;
  for (long long i = (long long)0; i < (long long)n; i++) cin >> a[i];
  for (long long i = (long long)0; i < (long long)n; i++) st.insert(a[i]);
  if (st.size() > k) {
    cout << "-1\n";
    return;
  }
  cout << n * k << "\n";
  for (long long i = (long long)0; i < (long long)n; i++) {
    for (long long var : st) cout << var << " ";
    for (long long j = 0; j < k - (long long)(st.size()); j++) cout << "1 ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solve();
}