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>
void build_comp(int v, int comp_num, std::vector<int> &cur_comp,
                std::vector<int> &rev_comps,
                const std::vector<std::vector<int>> &g) {
  cur_comp.push_back(v);
  rev_comps[v] = comp_num;
  for (int to : g[v]) {
    if (rev_comps[to] == -1) {
      build_comp(to, comp_num, cur_comp, rev_comps, g);
    }
  }
}
void solve() {
  int n;
  std::cin >> n;
  std::vector<std::vector<std::pair<int, int>>> positions(n);
  const auto reader = [n, &positions](int num) {
    for (int i = 0; i < n; ++i) {
      int a;
      std::cin >> a;
      positions[a - 1].emplace_back(num, i);
    }
  };
  const auto neg = [n](int val) { return (val + n) % (2 * n); };
  reader(0);
  reader(1);
  for (int i = 0; i < n; ++i) {
    if (positions[i].size() != 2) {
      std::cout << -1 << "\n";
      return;
    }
  }
  std::vector<std::vector<int>> g(2 * n);
  for (int i = 0; i < n; ++i) {
    auto pa = positions[i][0];
    auto pb = positions[i][1];
    if (pa.second == pb.second) {
      continue;
    }
    int a = pa.second;
    int b = pb.second;
    if (pa.first == pb.first) {
      a = neg(a);
    }
    g[a].push_back(b);
    g[b].push_back(a);
    g[neg(a)].push_back(neg(b));
    g[neg(b)].push_back(neg(a));
  }
  int comp_num = 0;
  std::vector<std::vector<int>> comps;
  std::vector<int> rev_comps(2 * n, -1);
  for (int i = 0; i < 2 * n; ++i) {
    if (rev_comps[i] == -1) {
      std::vector<int> cur_comp;
      build_comp(i, comp_num++, cur_comp, rev_comps, g);
      comps.emplace_back(std::move(cur_comp));
    }
  }
  for (int i = 0; i < n; ++i) {
    if (rev_comps[i] == rev_comps[neg(i)]) {
      std::cout << -1 << "\n";
      return;
    }
  }
  std::set<int> res;
  std::set<size_t> skipping_comps;
  for (size_t i = 0; i < comps.size(); ++i) {
    if (skipping_comps.find(i) != skipping_comps.end()) {
      continue;
    }
    const auto &line = comps[i];
    int positive = 0;
    for (int v : line) {
      if (v < n) {
        ++positive;
      }
    }
    int negative = line.size() - positive;
    if (positive <= negative) {
      for (int v : line) {
        if (v < n) {
          res.insert(v);
        }
      }
    }
    if (positive == negative) {
      skipping_comps.insert(rev_comps[neg(line[0])]);
    }
  }
  std::cout << res.size() << "\n";
  for (int v : res) {
    std::cout << v + 1 << " ";
  }
  std::cout << "\n";
}
int main() {
  int t;
  std::cin >> t;
  for (; t; --t) {
    solve();
  }
  return 0;
}
