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;
const long long mod = 1000000007;
const long long mod2 = 1ll << 32;
bool compare(pair<int, int> quan1, pair<int, int> quan2) {
  return quan1.second < quan2.second;
}
long long power(long long base, int expo) {
  long long ans = 1;
  while (expo > 0) {
    if (expo % 2 == 1) {
      ans *= base;
    }
    base *= base;
    expo /= 2;
  }
  return ans;
}
long long minimax(long long t, bool isMax, long long score, long long h,
                  long long scores[]) {
  if (t + 1 == h) return score;
  if (isMax && scores[t] != 1) {
    if (scores[t] % 2 == 0) {
      return max(minimax(t + 1, false, score + scores[t] - 1, h, scores),
                 minimax(t + 1, true, score + scores[t], h, scores));
    } else {
      return max(minimax(t + 1, true, score + scores[t] - 1, h, scores),
                 minimax(t + 1, false, score + scores[t], h, scores));
    }
  } else if (isMax && scores[t] == 1) {
    return minimax(t + 1, false, score + 1, h, scores);
  } else if (!isMax && scores[t] != 1) {
    if (scores[t] % 2 == 0) {
      return max(minimax(t + 1, true, score, h, scores),
                 minimax(t + 1, false, score, h, scores));
    } else {
      return max(minimax(t + 1, false, score, h, scores),
                 minimax(t + 1, true, score, h, scores));
    }
  } else if (!isMax && scores[t] == 1) {
    return minimax(t + 1, true, score, h, scores);
  }
}
long long func(long long arr[], long long n, long long k) {
  int calculate[n][31];
  for (auto i = 0; i < n; i++) {
    for (auto j = 0; j <= 30; j++) {
      calculate[i][j] = 0;
    }
  }
  long long answer = 0;
  vector<pair<long long, int>> netvalue;
  long long y = 0;
  for (auto i = 0; i < n; i++) {
    while (arr[i] > 0) {
      calculate[i][y] = arr[i] % 2;
      arr[i] = arr[i] / 2;
      y++;
    }
    reverse(calculate[i], calculate[i] + 31);
    y = 0;
  }
  for (auto j = 0; j <= 30; j++) {
    for (auto i = 1; i < n; i++) {
      calculate[i][j] += calculate[i - 1][j];
    }
  }
  for (auto i = 0; i <= 30; i++) {
    netvalue.push_back(
        make_pair((power(2, (30 - i)) * calculate[n - 1][i]), (30 - i)));
  }
  sort(netvalue.begin(), netvalue.end(), compare);
  long long u = 0;
  while (k > 0) {
    answer += power(2, netvalue[u].second);
    k--;
    u++;
  }
  return answer;
}
void dfSearch(int person, int visited[], unordered_set<int> &settt,
              vector<pair<int, long double>> vec[],
              unordered_map<int, long double> personTotime) {
  visited[person] = 1;
  long double lasttime;
  lasttime = personTotime[person];
  settt.insert(person);
  for (int j = 0; j < vec[person].size(); j++) {
    if ((visited[vec[person][j].first] == 0) &&
        (vec[person][j].second >= lasttime)) {
      personTotime[vec[person][j].first] = vec[person][j].second;
      dfSearch(vec[person][j].first, visited, settt, vec, personTotime);
    }
  }
}
long long lessFindit(long long ele) {
  long long start = 1;
  long long ending = 100000000;
  long long middle;
  long long ans = 0;
  while (start <= ending) {
    middle = (start + ending) / 2;
    if (ele >= ((middle * (middle + 1)) / 2)) {
      start = middle + 1;
      ans = middle;
    } else {
      ending = middle - 1;
    }
  }
  return ans;
}
long long exponentiation(long long num_one, long long num_two, long long mod) {
  long long answer = 1;
  num_one = num_one % mod;
  while (num_two > 0) {
    if (num_two & 1) {
      answer = (answer * num_one) % mod;
    }
    num_two = num_two >> 1;
    num_one = (num_one * num_one) % mod;
  }
  return answer % mod;
}
long long mod_inverse(long long number, long long mod) {
  return exponentiation(number, mod - 2, mod) % mod;
}
long long combi(long long n, long long k, long long mod) {
  if (k == 0) {
    return 1;
  }
  long long factorial[n + 1];
  factorial[0] = 1;
  for (long long i = 1; i <= n; i++) {
    factorial[i] = factorial[i - 1] * i % mod;
  }
  return ((factorial[n] % mod) * mod_inverse(factorial[k], mod) % mod *
          mod_inverse(factorial[n - k], mod) % mod) %
         mod;
}
long long fastModularExpo(long long base, long long expo) {
  long long answer = 1;
  base = base % mod;
  while (expo > 0) {
    if (expo % 2 == 1) {
      answer *= base;
      answer = answer % mod;
    }
    base *= base;
    base = base % mod;
    expo /= 2;
  }
  return (answer % mod);
}
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
long w;
void fun(vector<pair<long, long>> vec, long n, long max_cost[100][101]) {
  for (long i = 0; i < n; i++) {
    for (long j = 1; j <= w; j++) {
      if (vec[i].second <= j) {
        max_cost[i][j] =
            max(max_cost[i][j], max_cost[i][j - vec[i].second] + vec[i].first);
      } else {
        max_cost[i][j] = max_cost[i - 1][j];
      }
    }
  }
}
long calculate(vector<long> v, long l, long r, long k) {
  v.push_back(l);
  v.push_back(r);
  k = k / 2;
  long sum = 0;
  for (long i = 0; i <= (v.size() - 1 - k); i++) {
    sum += v[i];
  }
  return sum;
}
long long minval(long long val[], long long included[], long long n) {
  long long minim = INT_MAX;
  long long vertex;
  for (long long i = 0; i < n; i++) {
    if (val[i] < minim && included[i] == 0) {
      minim = val[i];
      vertex = i;
    }
  }
  return vertex;
}
void getmincost(map<pair<long long, long long>, long long> visited,
                long long sx, long long sy, long long fx, long long fy,
                long long &mini_cost, long long cost) {
  if (fx == sx && fy == sy) {
    mini_cost = min(mini_cost, cost);
    return;
  }
  visited[{sx, sy}] = 1;
  if ((((sy + (2 * sx)) < 0) && visited[{(-sx), (-(sy + (2 * sx)))}] == 0) &&
      cost <= 5) {
    getmincost(visited, -sx, -(sy + (2 * sx)), fx, fy, mini_cost, cost + 1);
  }
  if ((((sy + (2 * sx)) > 0) && visited[{sx, (sy + (2 * sx))}] == 0) &&
      cost <= 5) {
    getmincost(visited, sx, sy + (2 * sx), fx, fy, mini_cost, cost + 1);
  }
  if ((((sy - (2 * sx)) > 0) && visited[{sx, (sy - (2 * sx))}] == 0) &&
      cost <= 5) {
    getmincost(visited, sx, sy - (2 * sx), fx, fy, mini_cost, cost + 1);
  }
  if ((((sy - (2 * sx)) < 0) && visited[{(-sx), (-(sy - (2 * sx)))}] == 0) &&
      cost <= 5) {
    getmincost(visited, -sx, -(sy - (2 * sx)), fx, fy, mini_cost, cost + 1);
  }
  if ((visited[{(sx + (2 * sy)), sy}] == 0) && cost <= 5) {
    getmincost(visited, sx + (2 * sy), sy, fx, fy, mini_cost, cost + 1);
  }
  if ((visited[{(sx - (2 * sy)), sy}] == 0) && cost <= 5) {
    getmincost(visited, sx - (2 * sy), sy, fx, fy, mini_cost, cost + 1);
  }
  visited[{sx, sy}] = 0;
}
long long LCM(long long aa[], long long n) {
  long long lcm = aa[0];
  for (long long i = 1; i < n; i++) {
    lcm = (((aa[i] * lcm)) / (gcd(aa[i], lcm)));
  }
  return lcm;
}
long long mxcuts(long long cuts, vector<long long> v, long long limit,
                 long long i, long long n) {
  if (limit < 0) {
    return cuts - 1;
  }
  if (i == n || limit == 0) {
    return cuts;
  }
  return max(mxcuts(cuts + 1, v, limit - v[i], i + 1, n),
             mxcuts(cuts, v, limit, i + 1, n));
}
long long factori(long long n) {
  long long factp[n + 1];
  factp[0] = 1;
  factp[1] = 1;
  for (long long i = 1; i <= n; i++) {
    factp[i] = ((factp[i - 1] * i) + mod) % mod;
  }
  return factp[n];
}
long long findparent(long long u,
                     unordered_map<long long, vector<long long>> &parent,
                     long long p[]) {
  long long cpy = u;
  while (p[cpy] != cpy) {
    cpy = p[cpy];
  }
  p[u] = cpy;
  return cpy;
}
void findd(unordered_map<long long, long long> visited, long long &c,
           long long &maxi, long long &mini, long long u,
           unordered_map<long long, vector<long long>> &parent, long long p[]) {
  visited[u] = 1;
  c++;
  mini = min(mini, u);
  maxi = max(maxi, u);
  for (long long i = 0; i < parent[u].size(); i++) {
    if (visited[parent[u][i]] == 0) {
      findd(visited, c, maxi, mini, parent[u][i], parent, p);
    }
  }
}
vector<long long> finding_all_prime_factors(long long n) {
  vector<long long> v;
  while (n % 2 == 0) {
    v.push_back(2);
    n = n / 2;
  }
  for (long long y = 3; y * y <= (n); y += 2) {
    while (n % y == 0) {
      v.push_back(y);
      n = n / y;
    }
  }
  if (n > 2) {
    v.push_back(n);
  }
  return v;
}
void uni(long long u, long long v,
         unordered_map<long long, vector<long long>> &parent, long long p[]) {
  long long par1 = findparent(u, parent, p);
  long long par2 = findparent(v, parent, p);
  if (par1 != par2) {
    parent[par1].push_back(par2);
    parent[par2].push_back(par1);
    p[par1] = par2;
  }
}
long long dfs(unordered_map<long long, long long> &size, long long node,
              bool visit[], vector<long long> adj[]) {
  visit[node] = true;
  long long s = 1;
  for (int i = 0; i < adj[node].size(); i++) {
    if (visit[adj[node][i]] == false) {
      s += dfs(size, adj[node][i], visit, adj);
    }
  }
  size[node] = s;
  return size[node];
}
long long node;
void b_f_s(long long visited[], queue<long long> q, long long vertex,
           vector<long long> v[], vector<long long> adj[]) {
  q.push(vertex);
  visited[vertex] = 1;
  while (q.size() > 0) {
    node = q.front();
    visited[node] = 1;
    q.pop();
    for (long long i = 0; i < adj[node].size(); i++) {
      if (visited[adj[node][i]] == 0) {
        q.push(adj[node][i]);
        v[adj[node][i]] = v[node];
        v[adj[node][i]].push_back(node);
      }
    }
  }
}
long long optimizer(long long t, bool ischef, long long score, long long h,
                    long long scores[], long long n) {
  if (h == 0) {
    return score;
  }
  if (t == n) {
    t -= n;
  }
  if (ischef && scores[t] != 1) {
    if (t < n - 1) {
      if (scores[t] % 2 == 0) {
        return max(optimizer(t + 1, false, (score + scores[t] - 1) % mod, h - 1,
                             scores, n),
                   optimizer(t + 1, true, (score + scores[t]) % mod, h - 1,
                             scores, n));
      } else {
        return max(optimizer(t + 1, true, (score + scores[t] - 1) % mod, h - 1,
                             scores, n),
                   optimizer(t + 1, false, (score + scores[t]) % mod, h - 1,
                             scores, n));
      }
    } else {
      if (scores[t] % 2 == 0) {
        return max(optimizer(t + 1, true, (score + scores[t] - 1) % mod, h - 1,
                             scores, n),
                   optimizer(t + 1, false, (score + scores[t]) % mod, h - 1,
                             scores, n));
      } else {
        return max(optimizer(t + 1, false, (score + scores[t] - 1) % mod, h - 1,
                             scores, n),
                   optimizer(t + 1, true, (score + scores[t]) % mod, h - 1,
                             scores, n));
      }
    }
  } else if (ischef && scores[t] == 1) {
    if (t < n - 1) {
      return optimizer(t + 1, false, (score + 1) % mod, h - 1, scores, n);
    } else {
      return optimizer(t + 1, true, (score + 1) % mod, h - 1, scores, n);
    }
  } else if (!ischef && scores[t] != 1) {
    if (t < n - 1) {
      if (scores[t] % 2 == 0) {
        return max(optimizer(t + 1, true, score, h - 1, scores, n),
                   optimizer(t + 1, false, score, h - 1, scores, n));
      } else {
        return max(optimizer(t + 1, false, score, h - 1, scores, n),
                   optimizer(t + 1, true, score, h - 1, scores, n));
      }
    } else {
      if (scores[t] % 2 == 0) {
        return max(optimizer(t + 1, false, score, h - 1, scores, n),
                   optimizer(t + 1, true, score, h - 1, scores, n));
      } else {
        return max(optimizer(t + 1, true, score, h - 1, scores, n),
                   optimizer(t + 1, false, score, h - 1, scores, n));
      }
    }
  } else if (!ischef && scores[t] == 1) {
    if (t < n - 1) {
      return optimizer(t + 1, true, score, h - 1, scores, n);
    } else {
      return optimizer(t + 1, false, score, h - 1, scores, n);
    }
  }
}
long long sieve_to_count_primes(long long n) {
  long long is_it_prime[1000001];
  for (long long i = 1; i <= 1000000; i++) {
    is_it_prime[i] = 1;
  }
  for (long long i = 2; i <= sqrt(1000000); i++) {
    if (is_it_prime[i] == 1) {
      for (long long j = i * i; j <= 1000000;) {
        is_it_prime[j] = 0;
        j += i;
      }
    }
  }
}
long long LIS_Size(vector<long long> &v) {
  if (v.size() == 0) {
    return 0;
  }
  vector<long long> vv(v.size(), 0);
  long long size = 1;
  vv[0] = v[0];
  for (long long i = 1; i < v.size(); i++) {
    auto b = vv.begin(), e = vv.begin() + size;
    auto iter = lower_bound(b, e, v[i]);
    if (iter == vv.begin() + size) {
      vv[size] = v[i];
      size++;
    } else {
      *iter = v[i];
    }
  }
  return size;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t, n, quer, a, b, q, d, m, k, pos;
  long long x;
  long long num;
  long long last;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    long long aa[n + 1];
    vector<long long> v;
    unordered_set<long long> s;
    for (long long i = 0; i < n; i++) {
      cin >> aa[i];
      v.push_back(aa[i]);
      s.insert(aa[i]);
    }
    long long nn = n;
    long long f = 0;
    if (s.size() > k) {
      f = 1;
    }
    if (f == 1) {
      cout << -1 << endl;
    } else {
      cout << nn * k << endl;
      long long tt = nn;
      while (tt) {
        for (auto i : s) {
          cout << i << " ";
          last = i;
        }
        for (long long i = 0; i < (k - s.size()); i++) {
          cout << last << " ";
        }
        tt--;
      }
      cout << endl;
    }
  }
  return 0;
}