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;
char local = 'O';
template <typename T>
void read(T &a) {
  cin >> a;
}
template <typename T, typename T1>
void read(T &a, T1 &b) {
  cin >> a >> b;
}
template <typename T, typename T1, typename T2>
void read(T &a, T1 &b, T2 &c) {
  cin >> a >> b >> c;
}
template <typename T, typename T1>
void read(pair<T, T1> &p) {
  cin >> p.first >> p.second;
}
template <typename T>
void read(T arr[], long long int n) {
  for (long long int i = 0; i < n; i++) {
    read(arr[i]);
  }
}
template <typename T>
void read(vector<T> &arr) {
  for (long long int i = 0; i < arr.size(); i++) {
    read(arr[i]);
  }
}
long long int modularExponentiation(long long int x, long long int n,
                                    long long int M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modularExponentiation((x * x) % M, n / 2, M);
  return (x * modularExponentiation((x * x) % M, (n - 1) / 2, M)) % M;
}
long long int binaryExponentiation(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return binaryExponentiation(x * x, n / 2);
  else
    return x * binaryExponentiation(x * x, (n - 1) / 2);
}
long long int GCD(long long int A, long long int B) {
  if (B == 0)
    return A;
  else
    return GCD(B, A % B);
}
long long int extended_GCD(long long int a, long long int b, long long int &x,
                           long long int &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = extended_GCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
long long int modinv(long long int a, long long int mod) {
  long long int x, y;
  extended_GCD(a, mod, x, y);
  if (x < 0) x += mod;
  return x;
}
long long int modPrimeInverse(long long int A, long long int M) {
  return modularExponentiation(A, M - 2, M);
}
void sieve(long long int N) {
  bool isPrime[N + !1];
  for (long long int i = 0; i <= N; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (long long int i = 2; i * i <= N; ++i) {
    if (isPrime[i] == true) {
      for (long long int j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
}
long long int minPrime[1000007];
void factorSieve(long long int n) {
  memset(minPrime, 0, sizeof(minPrime));
  for (long long int i = 2; i * i <= n; ++i) {
    if (minPrime[i] == 0) {
      for (long long int j = i * i; j <= n; j += i) {
        if (minPrime[j] == 0) {
          minPrime[j] = i;
        }
      }
    }
  }
  for (long long int i = 2; i <= n; ++i) {
    if (minPrime[i] == 0) {
      minPrime[i] = i;
    }
  }
}
long long int factorize(long long int n) {
  map<long long int, long long int> m;
  m.clear();
  while (n != 1) {
    m[minPrime[n]]++;
    n /= minPrime[n];
  }
  for (auto i : m) {
    if (i.second > 1) {
      return 0;
    }
  }
  return 1;
}
void merge(long long int A[], long long int start, long long int mid,
           long long int end) {
  long long int p = start, q = mid + 1;
  long long int Arr[end - start + 1], k = 0;
  for (long long int i = start; i <= end; i++) {
    if (p > mid)
      Arr[k++] = A[q++];
    else if (q > end)
      Arr[k++] = A[p++];
    else if (A[p] < A[q])
      Arr[k++] = A[p++];
    else
      Arr[k++] = A[q++];
  }
  for (long long int p = 0; p < k; p++) {
    A[start++] = Arr[p];
  }
}
void merge_sort(long long int A[], long long int start, long long int end) {
  if (start < end) {
    long long int mid = (start + end) / 2;
    merge_sort(A, start, mid);
    merge_sort(A, mid + 1, end);
    merge(A, start, mid, end);
  }
}
long long int binarySearch(long long int a[], long long int low,
                           long long int high, long long int key) {
  while (low <= high) {
    long long int mid = (low + high) / 2;
    if (a[mid] < key) {
      low = mid + 1;
    } else if (a[mid] > key) {
      high = mid - 1;
    } else {
      return mid;
    }
  }
  return -1;
}
bool sortbysec(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  if (a.first == b.first) {
    a.second > b.second;
  }
  return (a.first < b.first);
}
void fib(long long int n, long long int m, long long int a[]) {
  if (n == 0) {
    a[0] = 0;
    a[1] = 1;
    return;
  }
  fib(n / 2, m, a);
  long long int c, d;
  c = (a[0] * (2 * a[1] - a[0] + m)) % m;
  d = (a[0] * a[0] + a[1] * a[1]) % m;
  if (n % 2 == 0) {
    a[0] = c;
    a[1] = d;
  } else {
    a[0] = d;
    a[1] = c + d;
  }
}
long long int st[2005][10];
vector<long long int> ca[9];
long long int c[20];
long long int n, k;
long long int dp[2005][2005];
string ans = "";
long long int check(long long int x, long long int r) {
  if (x == n) {
    if (r == 0) {
      return 0;
    } else {
      return 1;
    }
  }
  if (r < 0) {
    return 1;
  }
  if (dp[x][r] != 0) {
    return 1;
  }
  for (long long int i = 9; i >= 0; i--) {
    if (st[x][i] == -1 || st[x][i] > r) {
      continue;
    }
    if (check(x + 1, r - st[x][i]) == 0) {
      ans += '0' + i;
      return 0;
    }
  }
  return dp[x][r] = 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    long long int a[n];
    map<long long int, long long int> m;
    vector<long long int> b;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      if (m[a[i]] == 0) {
        m[a[i]] = 1;
        b.push_back(a[i]);
      }
    }
    if (b.size() > k) {
      cout << -1 << "\n";
      continue;
    }
    while (b.size() != k) {
      b.push_back(b[0]);
    }
    vector<long long int> ans;
    long long int flag = 0;
    long long int j = 0;
    for (long long int i = 0; i < n; i++) {
      long long int c = 0;
      j %= k;
      while (c < k && a[i] != b[j % k]) {
        ans.push_back(b[j % k]);
        j++;
        j %= k;
        c++;
      }
      if (c == k) {
        flag = 1;
        break;
      } else {
        ans.push_back(b[j]);
        j++;
        j %= k;
      }
    }
    if (flag == 1) {
      cout << -1 << "\n";
    } else {
      cout << ans.size() << "\n";
      for (auto i : ans) {
        cout << i << " ";
      }
      cout << "\n";
    }
  }
}