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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
bool MinPlace(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool MaxPlace(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename S, typename T>
ostream &operator<<(ostream &out, const pair<S, T> &p) {
  out << "{" << p.first << ", " << p.second << "}";
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  out << "[";
  for (int i = 0; i < (int)v.size(); i++) {
    out << v[i];
    if (i != (int)v.size() - 1) out << ", ";
  }
  out << "]";
  return out;
}
void dfs(int u, vector<bool> &vis, int ch, vector<int> &stay, vector<int> &flip,
         vector<vector<pair<int, int>>> &gr) {
  vis[u] = true;
  if (ch)
    flip.push_back(u);
  else
    stay.push_back(u);
  for (auto vv : gr[u]) {
    int v = vv.first, type = vv.second ^ ch;
    if (!vis[v]) dfs(v, vis, type, stay, flip, gr);
  }
}
void solve() {
  int n;
  cin >> n;
  vector<vector<int>> a(2, vector<int>(n));
  vector<vector<pair<int, int>>> gr(n, vector<pair<int, int>>());
  vector<vector<pair<int, int>>> where(n, vector<pair<int, int>>());
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < n; j++) {
      cin >> a[i][j];
      where[--a[i][j]].push_back(make_pair(i, j));
    }
  for (int i = 0; i < n; i++) {
    if (where[i].size() != 2) {
      cout << "-1\n";
      return;
    }
    int u = where[i][0].second, v = where[i][1].second;
    int type = (where[i][0].first == where[i][1].first);
    gr[u].push_back({v, type});
    gr[v].push_back({u, type});
  }
  vector<int> out;
  vector<bool> vis(n, false);
  for (int i = 0; i < n; i++)
    if (!vis[i]) {
      vector<int> flip, stay;
      dfs(i, vis, false, stay, flip, gr);
      if (stay.size() < flip.size()) swap(stay, flip);
      out.insert(out.end(), flip.begin(), flip.end());
    }
  cout << out.size() << "\n";
  for (auto idx : out) cout << (idx + 1) << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) solve();
};
