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;
int main(int argc, char** argv) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(15);
  if (argc == 2 && atoi(argv[1]) == 123456789)
    freopen("d:\\code\\cpp\\contests\\stdin", "r", stdin);
  int T;
  cin >> T;
  for (int(tt) = 0; (tt) < (T); (tt)++) {
    int n;
    cin >> n;
    vector<vector<int>> a(2, vector<int>(n + 1));
    vector<vector<int>> pos(n + 1);
    for (int(i) = 0; (i) < (2); (i)++) {
      for (int(j) = 1; (j) <= (n); (j)++) {
        int x;
        cin >> x;
        a[i][j] = x;
        pos[x].push_back(i == 0 ? j : -j);
      }
    }
    vector<bool> done(n + 1);
    vector<vector<int>> edges(n + 1);
    bool ok = 1;
    for (int(i) = 1; (i) <= (n); (i)++) {
      if (pos[i].size() != 2) {
        ok = 0;
        break;
      } else {
        int u = abs(pos[i][0]);
        int v = abs(pos[i][1]);
        if (u != v) {
          edges[u].push_back(v);
          edges[v].push_back(u);
        } else {
          done[u] = 1;
        }
      }
    }
    vector<int> sol;
    for (int(i) = 1; (i) <= (n); (i)++) {
      if (!done[i]) {
        int u = i;
        done[u] = 1;
        vector<vector<int>> t(2);
        int id = 1;
        t[id].push_back(u);
        do {
          for (auto v : edges[u])
            if (!done[v]) {
              if (a[0][u] == a[0][v] || a[1][u] == a[1][v]) id ^= 1;
              u = v;
              t[id].push_back(u);
              break;
            }
          if (done[u]) break;
          done[u] = 1;
        } while (u != i);
        if (t[0].size() > t[1].size()) swap(t[0], t[1]);
        sol.insert(sol.end(), t[0].begin(), t[0].end());
      }
    }
    if (!ok)
      cout << -1 << endl;
    else {
      cout << sol.size() << endl;
      for (auto it : sol) cout << it << " ";
      cout << endl;
    }
  }
  if (argc == 2 && atoi(argv[1]) == 123456789)
    cout << clock() * 1.0 / CLOCKS_PER_SEC << " sec\n";
  return 0;
}
