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 <typename Arg1>
void prn(Arg1&& arg1) {
  cout << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void prn(Arg1&& arg1, Args&&... args) {
  cout << arg1 << " ";
  prn(args...);
}
template <typename Arg1>
void prs(Arg1&& arg1) {
  cout << arg1 << " ";
}
template <typename Arg1, typename... Args>
void prs(Arg1&& arg1, Args&&... args) {
  cout << arg1 << " ";
  prs(args...);
}
template <typename Arg1>
void read(Arg1&& arg1) {
  cin >> arg1;
}
template <typename Arg1, typename... Args>
void read(Arg1&& arg1, Args&&... args) {
  cin >> arg1;
  read(args...);
}
int n;
vector<vector<int> > pos, adj;
vector<int> ogCol;
vector<bool> seen[2];
vector<int> changes[2];
void dfs(int cur, int col, int sid) {
  seen[sid][cur] = 1;
  if (col != ogCol[cur]) {
    if (cur < n)
      changes[sid].push_back(cur + 1);
    else
      changes[sid].push_back(cur - n + 1);
  }
  col ^= 1;
  for (auto i : adj[cur]) {
    if (!seen[sid][i]) dfs(i, col, sid);
  }
}
void add(int a, int b) {
  adj[a].push_back(b);
  adj[b].push_back(a);
}
void solve() {
  read(n);
  pos.assign(n + 1, vector<int>());
  ogCol.resize(2 * n);
  adj.assign(2 * n, vector<int>());
  for (int i = 0; i < n; ++i) {
    ogCol[i] = 0;
    int x;
    read(x);
    if (!pos[x].empty()) {
      add(i, pos[x][0]);
    }
    pos[x].push_back(i);
  }
  for (int i = 0; i < n; ++i) {
    add(n + i, i);
    ogCol[i + n] = 1;
    int x;
    read(x);
    if (!pos[x].empty()) {
      add(n + i, pos[x][0]);
    }
    pos[x].push_back(n + i);
  }
  for (int i = 1; i <= n; ++i) {
    if (pos[i].size() != 2) {
      prn(-1);
      return;
    }
  }
  for (int i = 0; i < 2; ++i) seen[i].assign(2 * n, 0);
  vector<int> ans;
  for (int i = 0; i < 2 * n; ++i) {
    if (seen[0][i]) continue;
    for (int j = 0; j < 2; ++j) changes[j].clear();
    for (int j = 0; j < 2; ++j) dfs(i, j, j);
    if (changes[0].size() < changes[1].size())
      ans.insert(ans.end(), changes[0].begin(), changes[0].end());
    else
      ans.insert(ans.end(), changes[1].begin(), changes[1].end());
  }
  sort(ans.begin(), ans.end());
  ans.erase(unique(ans.begin(), ans.end()), ans.end());
  prn(ans.size());
  for (auto i : ans) {
    prs(i);
  }
  prn("");
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  read(t);
  for (int tc = 1; tc <= t; ++tc) {
    solve();
  }
  return 0;
}
