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;
using ll = long long;
using ull = unsigned long long;
bool valid(const vector<int> &ais, const vector<int> &bis) {
  int n = ais.size();
  vector<int> count(n, 0);
  for (int i = 0; i < n; i += 1) {
    count[ais[i]]++;
    count[bis[i]]++;
  }
  for (int i = 0; i < n; i += 1) {
    if (count[i] != 2) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  for (int _case = 0; _case < t; _case += 1) {
    int n;
    cin >> n;
    vector<int> ais(n), bis(n);
    for (int i = 0; i < n; i += 1) {
      cin >> ais[i];
      ais[i]--;
    }
    for (int i = 0; i < n; i += 1) {
      cin >> bis[i];
      bis[i]--;
    }
    if (!valid(ais, bis)) {
      cout << -1 << endl;
      continue;
    }
    vector<bool> used(2 * n, false);
    vector<vector<pair<int, int>>> graph(2 * n);
    for (int column = 0; column < n; column += 1) {
      graph[column].emplace_back(ais[column] + n, 0);
      graph[ais[column] + n].emplace_back(column, 0);
      graph[column].emplace_back(bis[column] + n, 1);
      graph[bis[column] + n].emplace_back(column, 1);
    }
    vector<int> sol;
    for (int column = 0; column < n; column += 1) {
      if (used[column]) continue;
      vector<vector<int>> vecs(2);
      int curr = column, c = 0;
      while (true) {
        if (used[curr]) break;
        used[curr] = true;
        int n1 = graph[curr][0].first, n2 = graph[curr][1].first;
        int w1 = graph[curr][0].second, w2 = graph[curr][1].second;
        if (used[n1] && used[n2]) {
          if (n1 == column) {
            if (w1 == 1) vecs[c].push_back(column);
          } else {
            if (w2 == 1) vecs[c].push_back(column);
          }
          curr = column;
        } else if (!used[n1]) {
          if (w1 == 1) vecs[c].push_back(curr < n ? curr : n1);
          curr = n1;
        } else {
          if (w2 == 1) vecs[c].push_back(curr < n ? curr : n2);
          curr = n2;
        }
        c = (c + 1) % 2;
      }
      if (vecs[c].size() > vecs[(c + 1) % 2].size()) c = (c + 1) % 2;
      for (const auto el : vecs[c]) sol.push_back(el + 1);
    }
    cout << sol.size() << endl;
    for (const auto el : sol) cout << " " << el;
    cout << endl;
  }
}
