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;
void dfs(int node, vector<int> &visited,
         vector<vector<pair<int, int> > > &graph, vector<int> &s,
         vector<int> &us, int x) {
  visited[node] = 1;
  if (x)
    s.push_back(node);
  else
    us.push_back(node);
  for (pair<int, int> adj : graph[node]) {
    if (!visited[adj.first]) {
      int t = x;
      if (adj.second) {
        t++;
        t %= 2;
      }
      dfs(adj.first, visited, graph, s, us, t);
    }
  }
}
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    for (int &x : a) cin >> x;
    for (int &x : b) cin >> x;
    vector<vector<pair<int, int> > > pos(n + 1);
    vector<int> cnt(n + 1);
    for (int i = 0; i < n; i++) {
      pos[a[i]].push_back({1, i});
      pos[b[i]].push_back({2, i});
      cnt[a[i]]++;
      cnt[b[i]]++;
    }
    bool bol = true;
    for (int i = 1; i <= n; i++) {
      if (cnt[i] != 2) bol = false;
    }
    if (bol) {
      vector<vector<pair<int, int> > > graph(n + 1);
      pos.erase(pos.begin(), pos.begin() + 1);
      for (vector<pair<int, int> > p : pos) {
        if (p[0].first == p[1].first) {
          graph[p[0].second].push_back({p[1].second, 1});
          graph[p[1].second].push_back({p[0].second, 1});
        } else {
          graph[p[0].second].push_back({p[1].second, 0});
          graph[p[1].second].push_back({p[0].second, 0});
        }
      }
      vector<int> ans;
      vector<int> visited(n);
      for (int i = 0; i < n; i++) {
        if (!visited[i]) {
          vector<int> s, us;
          dfs(i, visited, graph, s, us, 0);
          if (us.size() < s.size()) swap(s, us);
          ans.insert(ans.end(), s.begin(), s.end());
        }
      }
      cout << ans.size() << '\n';
      for (int x : ans) cout << x + 1 << "  ";
      cout << '\n';
    } else
      cout << "-1\n";
  }
}
