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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t-- > 0) {
    int n, k;
    cin >> n;
    vector<int> r1(n), r2(n);
    for (int i = 0; i < n; i++) cin >> r1[i];
    for (int i = 0; i < n; i++) cin >> r2[i];
    vector<vector<int>> pos(n + 1);
    for (int i = 0; i < n; i++) {
      pos[r1[i]].push_back(i);
      pos[r2[i]].push_back(i);
    }
    vector<vector<pair<int, int>>> edges(n + 1);
    bool flag = true;
    for (int i = 1; i < n + 1; i++) {
      if (pos[i].size() != 2) {
        flag = false;
        break;
      }
      if (pos[i][0] == pos[i][1]) continue;
      if (r1[pos[i][0]] == i && r1[pos[i][1]] == i ||
          r2[pos[i][0]] == i && r2[pos[i][1]] == i) {
        edges[pos[i][0]].push_back(pair<int, int>(pos[i][1], 1));
        edges[pos[i][1]].push_back(pair<int, int>(pos[i][0], 1));
      } else {
        edges[pos[i][0]].push_back(pair<int, int>(pos[i][1], 0));
        edges[pos[i][1]].push_back(pair<int, int>(pos[i][0], 0));
      }
    }
    if (!flag) {
      cout << -1 << endl;
      continue;
    }
    vector<int> vis(n + 1, false);
    list<int> res;
    for (int i = 1; i < n + 1; i++) {
      if (vis[i]) continue;
      vector<vector<int>> color(2);
      color[0].push_back(i);
      list<pair<int, int>> que;
      que.push_back(pair<int, int>(i, 0));
      vis[i] = true;
      while (!que.empty()) {
        auto x = que.front();
        que.pop_front();
        for (auto e : edges[x.first]) {
          if (!vis[e.first]) {
            vis[e.first] = true;
            int c = x.second ^ e.second;
            que.push_back(pair<int, int>(e.first, c));
            color[c].push_back(e.first);
          }
        }
      }
      if (color[0].size() < color[1].size()) {
        for (int i : color[0]) res.push_back(i + 1);
      } else {
        for (int i : color[1]) res.push_back(i + 1);
      }
    }
    cout << res.size() << endl;
    for (int i : res) cout << i << " ";
    cout << endl;
  }
  return 0;
}
