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 append(vector<int> &a, vector<int> &b) {
  for (auto e : b) a.push_back(e);
}
int main() {
  int t;
  cin >> t;
  cin.tie(0);
  cin.sync_with_stdio(true);
  while (t--) {
    int n;
    cin >> n;
    int a[n + 1], b[n + 1];
    vector<int> v(n + 1, 0);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      v[a[i]]++;
    }
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      v[b[i]]++;
    }
    int poss = 1;
    for (int i = 1; i <= n && poss; i++) {
      if (v[i] != 2) poss = 0;
    }
    if (poss) {
      set<pair<int, int>> g[n + 1];
      for (int i = 0; i < n; i++) {
        if (a[i] != b[i]) {
          g[a[i]].insert(make_pair(b[i], i + 1));
          g[b[i]].insert(make_pair(a[i], -(i + 1)));
        }
      }
      vector<int> swp, vis(n + 1, 0), ans;
      for (int i = 1; i <= n; i++) {
        vector<int> fwd[2];
        int u = i;
        while (!vis[u]) {
          vis[u] = 1;
          for (auto e : g[u]) {
            if (!vis[e.first]) {
              fwd[(e.second < 0 ? 1 : 0)].push_back(abs(e.second));
              u = e.first;
              break;
            }
          }
        }
        if (u != i) {
          for (auto e : g[u]) {
            if (e.first == i) {
              fwd[(e.second < 0 ? 1 : 0)].push_back(abs(e.second));
              break;
            }
          }
        }
        if (fwd[0].size() < fwd[1].size())
          append(ans, fwd[0]);
        else
          append(ans, fwd[1]);
      }
      cout << ans.size() << endl;
      for (auto e : ans) cout << e << " ";
      cout << endl;
    } else {
      cout << -1 << endl;
    }
  }
  return 0;
}
