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;
using ll = long long;
using dd = double;
using pll = pair<ll, ll>;
using tll = tuple<ll, ll, ll>;
using vll = vector<ll>;
using vdd = vector<dd>;
using vpll = vector<pll>;
using vtll = vector<tll>;
using vvll = vector<vll>;
using vvdd = vector<vdd>;
using vvpll = vector<vpll>;
using vvtll = vector<vtll>;
using vvvll = vector<vvll>;
using vvvdd = vector<vvdd>;
using vvvpll = vector<vvpll>;
using vvvtll = vector<vvtll>;
using vvvvll = vector<vvvll>;
using vvvvdd = vector<vvvdd>;
using vvvvpll = vector<vvvpll>;
using vvvvtll = vector<vvvtll>;
constexpr ll INF = 1LL << 60;
constexpr dd EPS = 1e-11;
struct Fast {
  Fast() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(std::numeric_limits<double>::max_digits10);
  }
} fast;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline T MaxE(vector<T> &v, ll S, ll E) {
  T m = v[S];
  for (ll i = (S); i <= (E); i++) chmax(m, v[i]);
  return m;
}
template <class T>
inline T MinE(vector<T> &v, ll S, ll E) {
  T m = v[S];
  for (ll i = (S); i <= (E); i++) chmin(m, v[i]);
  return m;
}
template <class T>
inline T MaxE(vector<T> &v, ll N) {
  return MaxE(v, 0, N - 1);
}
template <class T>
inline T MinE(vector<T> &v, ll N) {
  return MinE(v, 0, N - 1);
}
template <class T>
inline T MaxE(vector<T> &v) {
  return MaxE(v, (ll)v.size());
}
template <class T>
inline T MinE(vector<T> &v) {
  return MinE(v, (ll)v.size());
}
template <class T>
inline ll MaxI(vector<T> &v, ll S, ll E) {
  ll m = S;
  for (ll i = (S); i <= (E); i++) {
    if (v[i] > v[m]) m = i;
  }
  return m;
}
template <class T>
inline ll MinI(vector<T> &v, ll S, ll E) {
  ll m = S;
  for (ll i = (S); i <= (E); i++) {
    if (v[i] < v[m]) m = i;
  }
  return m;
}
template <class T>
inline ll MaxI(vector<T> &v, ll N) {
  return MaxI(v, 0, N - 1);
}
template <class T>
inline ll MinI(vector<T> &v, ll N) {
  return MinI(v, 0, N - 1);
}
template <class T>
inline ll MaxI(vector<T> &v) {
  return MaxI(v, (ll)v.size());
}
template <class T>
inline ll MinI(vector<T> &v) {
  return MinI(v, (ll)v.size());
}
template <class T>
inline T Sum(vector<T> &v, ll S, ll E) {
  T s = v[S];
  for (ll i = (S + 1); i <= (E); i++) s += v[i];
  return s;
}
template <class T>
inline T Sum(vector<T> &v, ll N) {
  return Sum(v, 0, N - 1);
}
template <class T>
inline T Sum(vector<T> &v) {
  return Sum(v, v.size());
}
template <class T>
inline ll sz(T &v) {
  return (ll)v.size();
}
template <class T>
inline T POW(T a, ll n) {
  T r = 1;
  for (; n > 0; n >>= 1, a *= a) {
    if (n & 1) r *= a;
  }
  return r;
}
inline ll POW(int a, ll n) { return POW((ll)a, n); }
inline ll CEIL(ll a, ll b) { return (a + b - 1) / b; }
ll Gcd(ll a, ll b) { return (b == 0) ? a : Gcd(b, a % b); }
ll Lcm(ll a, ll b) { return a * b / Gcd(a, b); }
inline ll MSB(ll a) {
  for (ll o = 63, x = -1;;) {
    ll m = (o + x) / 2;
    if (a < (1LL << m))
      o = m;
    else
      x = m;
    if (o - x == 1) return x;
  }
}
inline ll Upper2N(ll s) { return 1LL << MSB(s) << 1; }
inline vll Bit2Idx(ll m) {
  vll v;
  for (ll i = 0; m; m >>= 1, i++)
    if (m & 1) v.push_back(i);
  return move(v);
}
inline ll BitNum(ll m) {
  for (ll c = 0;; m >>= 1) {
    c += m & 1;
    if (!m) return c;
  }
}
inline ll Bit(ll s, ll i) { return (s >> i) & 1; }
[[nodiscard]] inline ll BitOn(ll s, ll i) { return s | (1LL << i); }
[[nodiscard]] inline ll BitOff(ll s, ll i) { return s & ~(1LL << i); }
[[nodiscard]] inline ll BitCut(ll s, ll i) { return s & (1LL << ++i) - 1; }
inline vll str2num(string &s, ll N) {
  vll v(N);
  for (ll i = (0); i <= ((N)-1); i++) v[i] = (ll)(s[i] - '0');
  return move(v);
}
inline vll ALP2num(string &s, ll N) {
  vll v(N);
  for (ll i = (0); i <= ((N)-1); i++) v[i] = (ll)(s[i] - 'A');
  return move(v);
}
inline vll alp2num(string &s, ll N) {
  vll v(N);
  for (ll i = (0); i <= ((N)-1); i++) v[i] = (ll)(s[i] - 'a');
  return move(v);
}
template <class T>
inline ll UnderNumOf(vector<T> &v, T x) {
  return upper_bound(v.begin(), v.end(), x) - v.begin();
}
template <class T, class Pr>
inline ll UnderNumOf(vector<T> &v, T x, Pr pred) {
  return upper_bound(v.begin(), v.end(), x, pred) - v.begin();
}
template <class T>
inline ll OverNumOf(vector<T> &v, T x) {
  return (ll)v.size() - (lower_bound(v.begin(), v.end(), x) - v.begin());
}
template <class T, class Pr>
inline ll OverNumOf(vector<T> &v, T x, Pr pred) {
  return (ll)v.size() - (lower_bound(v.begin(), v.end(), x, pred) - v.begin());
}
template <class T = ll>
inline vector<T> cinv(ll N) {
  vector<T> v(N);
  for (ll i = (0); i <= ((N)-1); i++) cin >> v[i];
  return move(v);
}
template <class T = ll, class S = ll>
inline vector<pair<T, S>> cinv2(ll N) {
  vector<pair<T, S>> v(N);
  for (ll i = (0); i <= ((N)-1); i++) {
    cin >> v[i].first >> v[i].second;
  }
  return move(v);
}
template <class T = ll, class S = ll, class R = ll>
inline vector<tuple<T, S, R>> cinv3(ll N) {
  vector<tuple<T, S, R>> v(N);
  for (ll i = (0); i <= ((N)-1); i++) {
    cin >> get<0>(v[i]) >> get<1>(v[i]) >> get<2>(v[i]);
  }
  return move(v);
}
template <class T = ll, class S = ll, class R = ll, class Q = ll>
inline vector<tuple<T, S, R, Q>> cinv4(ll N) {
  vector<tuple<T, S, R, Q>> v(N);
  for (ll i = (0); i <= ((N)-1); i++) {
    cin >> get<0>(v[i]) >> get<1>(v[i]) >> get<2>(v[i]) >> get<3>(v[i]);
  }
  return move(v);
}
template <class T = ll>
inline vector<vector<T>> cinvv(ll N, ll M) {
  vector<vector<T>> vv(N);
  for (ll i = (0); i <= ((N)-1); i++) vv[i] = cinv(M);
  return move(vv);
}
template <class T>
inline void coutv(vector<T> &v, char deli = ' ') {
  ll N = (ll)v.size();
  for (ll i = (0); i <= ((N)-1); i++) {
    cout << v[i] << ((i == N - 1) ? '\n' : deli);
  }
}
template <class T>
void bye(T a) {
  cout << a << '\n';
  exit(0);
}
template <class T>
void Dump(vector<T> &v) {}
template <class T>
void Dump(vector<vector<T>> &v) {}
template <class T>
void Dump(vector<vector<vector<T>>> &v) {}
template <class T>
void Dump(vector<vector<vector<vector<T>>>> &v) {}
template <class T>
void Dump(vector<vector<vector<vector<vector<T>>>>> &v) {}
template <ll MOD>
struct mll_ {
  ll val;
  mll_(ll v = 0) : val(v % MOD) {
    if (val < 0) val += MOD;
  }
  mll_ operator-() const { return -val; }
  mll_ operator+(const mll_ &b) const { return val + b.val; }
  mll_ operator-(const mll_ &b) const { return val - b.val; }
  mll_ operator*(const mll_ &b) const { return val * b.val; }
  mll_ operator/(const mll_ &b) const { return mll_(*this) /= b; }
  mll_ operator+(ll b) const { return *this + mll_(b); }
  mll_ operator-(ll b) const { return *this - mll_(b); }
  mll_ operator*(ll b) const { return *this * mll_(b); }
  friend mll_ operator+(ll a, const mll_ &b) { return b + a; }
  friend mll_ operator-(ll a, const mll_ &b) { return -b + a; }
  friend mll_ operator*(ll a, const mll_ &b) { return b * a; }
  friend mll_ operator/(ll a, const mll_ &b) { return mll_(a) / b; }
  mll_ &operator+=(const mll_ &b) {
    val = (val + b.val) % MOD;
    return *this;
  }
  mll_ &operator-=(const mll_ &b) {
    val = (val + MOD - b.val) % MOD;
    return *this;
  }
  mll_ &operator*=(const mll_ &b) {
    val = (val * b.val) % MOD;
    return *this;
  }
  mll_ &operator/=(const mll_ &b) {
    ll c = b.val, d = MOD, u = 1, v = 0;
    while (d) {
      ll t = c / d;
      c -= t * d;
      swap(c, d);
      u -= t * v;
      swap(u, v);
    }
    val = val * u % MOD;
    if (val < 0) val += MOD;
    return *this;
  }
  mll_ &operator+=(ll b) { return *this += mll_(b); }
  mll_ &operator-=(ll b) { return *this -= mll_(b); }
  mll_ &operator*=(ll b) { return *this *= mll_(b); }
  mll_ &operator/=(ll b) { return *this /= mll_(b); }
  bool operator==(const mll_ &b) { return val == b.val; }
  bool operator!=(const mll_ &b) { return val != b.val; }
  bool operator==(ll b) { return *this == mll_(b); }
  bool operator!=(ll b) { return *this != mll_(b); }
  friend bool operator==(ll a, const mll_ &b) { return mll_(a) == b.val; }
  friend bool operator!=(ll a, const mll_ &b) { return mll_(a) != b.val; }
  friend ostream &operator<<(ostream &os, const mll_ &a) { return os << a.val; }
  friend istream &operator>>(istream &is, mll_ &a) { return is >> a.val; }
  static mll_ Combination(ll a, ll b) {
    chmin(b, a - b);
    if (b < 0) return mll_(0);
    mll_ c = 1;
    for (ll i = (0); i <= ((b)-1); i++) c *= a - i;
    for (ll i = (0); i <= ((b)-1); i++) c /= i + 1;
    return c;
  }
};
using mll = mll_<1000000007LL>;
using vmll = std::vector<mll>;
using vvmll = std::vector<vmll>;
using vvvmll = std::vector<vvmll>;
using vvvvmll = std::vector<vvvmll>;
using vvvvvmll = std::vector<vvvvmll>;
void solve() {
  ll N, K;
  cin >> N >> K;
  vector<ll> A = cinv<ll>(N);
  vll cnt(N + 1);
  for (ll i = (0); i <= (N - 1); i++) {
    if (cnt[A[i]] == 0) cnt[A[i]]++;
  }
  ll KK = Sum(cnt);
  if (KK > K) {
    cout << -1 << '\n';
    return;
  }
  vll base;
  for (ll i = (1); i <= (N); i++) {
    if (cnt[i] > 0) {
      base.push_back(i);
    }
  }
  ll NN = sz(base);
  for (ll i = (NN); i <= (K - 1); i++) {
    base.push_back(base[i % NN]);
  }
  vll ans;
  ll idx = 0;
  ll j = 0;
  while (true) {
    if (idx >= N) break;
    ll a = base[j];
    ans.push_back(a);
    if (a == A[idx]) idx++;
    j++;
    j %= K;
  }
  cout << sz(ans) << '\n';
  coutv(ans);
}
int main() {
  ll t;
  cin >> t;
  for (ll i = (0); i <= (t - 1); i++) solve();
  return 0;
}