You are given a table a of size 2 × n (i.e. two rows and n columns) consisting of integers from 1 to n.

In one move, you can choose some column j (1 ≤ j ≤ n) and swap values a_{1, j} and a_{2, j} in it. Each column can be chosen no more than once.

Your task is to find the minimum number of moves required to obtain permutations of size n in both first and second rows of the table or determine if it is impossible to do that.

You have to answer t independent test cases.

Recall that the permutation of size n is such an array of size n that contains each integer from 1 to n exactly once (the order of elements doesn't matter).

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of columns in the table. The second line of the test case contains n integers a_{1, 1}, a_{1, 2}, ..., a_{1, n} (1 ≤ a_{1, i} ≤ n), where a_{1, i} is the i-th element of the first row of the table. The third line of the test case contains n integers a_{2, 1}, a_{2, 2}, ..., a_{2, n} (1 ≤ a_{2, i} ≤ n), where a_{2, i} is the i-th element of the second row of the table.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case print the answer: -1 if it is impossible to obtain permutation of size n in both first and the second rows of the table, or one integer k in the first line, where k is the minimum number of moves required to obtain permutations in both rows, and k distinct integers pos_1, pos_2, ..., pos_k in the second line (1 ≤ pos_i ≤ n) in any order — indices of columns in which you need to swap values to obtain permutations in both rows. If there are several answers, you can print any.

Example

Input


6
4
1 2 3 4
2 3 1 4
5
5 3 5 1 4
1 2 3 2 4
3
1 2 1
3 3 2
4
1 2 2 1
3 4 3 4
4
4 3 1 4
3 2 2 1
3
1 1 2
3 2 2


Output


0

2
2 3 
1
1 
2
3 4 
2
3 4 
-1
#include <bits/stdc++.h>
using namespace std;
template <class T>
void dbs(string str, T t) {
  cerr << str << " : " << t << "\n";
}
template <class T, class... S>
void dbs(string str, T t, S... s) {
  long long idx = str.find(',');
  cerr << str.substr(0, idx) << " : " << t << ", ";
  dbs(str.substr(idx + 1), s...);
}
template <class T>
void prc(T a, T b) {
  cerr << "[";
  for (T i = a; i != b; ++i) {
    if (i != a) cerr << ", ";
    cerr << *i;
  }
  cerr << "]\n";
}
template <class T>
void prall(T a) {
  prc(a.begin(), a.end());
}
const int inf = 1000 * 1000 * 1000 + 9;
const long long infl = 1ll * inf * inf;
int MOD = 1000 * 1000 * 1000 + 7;
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
long long randrange(long long l, long long r) { return l + mt() % (r - l + 1); }
const int N = 2e5 + 5;
vector<int> row[N];
vector<int> col[N];
vector<pair<int, int> > adj[N];
int vis[N];
vector<int> cc[2];
int color[N];
void dfs(int u) {
  cc[color[u]].push_back(u);
  vis[u] = 1;
  for (pair<int, int> pp : adj[u]) {
    int v = pp.first, e = pp.second;
    if (vis[v] == 0) {
      if (e == 0) {
        color[v] = color[u];
        dfs(v);
      } else {
        color[v] = 1 - color[u];
        dfs(v);
      }
    }
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    row[i].clear(), col[i].clear(), adj[i].clear(), vis[i] = 0;
  bool pos = 1;
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < n; j++) {
      int a;
      cin >> a;
      a--;
      row[a].push_back(i);
      col[a].push_back(j);
      if (((int)row[a].size()) > 2) {
        pos = 0;
      }
    }
  }
  if (!pos) {
    cout << -1 << '\n';
    return;
  }
  for (int i = 0; i < n; i++) {
    if (col[i][0] != col[i][1]) {
      if (row[i][0] == row[i][1]) {
        adj[col[i][0]].push_back({col[i][1], 1});
        adj[col[i][1]].push_back({col[i][0], 1});
      } else {
        adj[col[i][0]].push_back({col[i][1], 0});
        adj[col[i][1]].push_back({col[i][0], 0});
      }
    }
  }
  int ans = 0;
  vector<int> position;
  for (int i = 0; i < n; i++) {
    if (vis[i] == 0) {
      color[i] = 0;
      cc[0].clear();
      cc[1].clear();
      dfs(i);
      int amt;
      vector<int> tt;
      if (((int)cc[0].size()) > ((int)cc[1].size())) {
        amt = ((int)cc[1].size());
        tt = cc[1];
      } else
        amt = ((int)cc[0].size()), tt = cc[0];
      ans += amt;
      for (int v : tt) position.push_back(v);
    }
  }
  cout << ans << '\n';
  for (int p : position) cout << p + 1 << " ";
  cout << '\n';
}
int main() {
  clock_t beg = clock();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int testcases = 1;
  cin >> testcases;
  cout << fixed << setprecision(13);
  int cs = 1;
  while (testcases--) {
    cs++;
    solve();
  }
  clock_t end = clock();
  fprintf(stderr, "%lf sec\n", (double)(end - beg) / (CLOCKS_PER_SEC));
}
