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_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t-- > 0) {
    int n;
    cin >> n;
    bool good = true;
    vector<int> a;
    vector<pair<int, int>> rs(n, {-1, -1});
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      x--;
      a.push_back(x);
      if (rs[x].first < 0)
        rs[x].first = i;
      else if (rs[x].second < 0)
        rs[x].second = i;
      else
        good = false;
    }
    vector<pair<int, int>> b;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      x--;
      b.push_back({a[i], x});
      if (rs[x].first < 0)
        rs[x].first = i;
      else if (rs[x].second < 0)
        rs[x].second = i;
      else
        good = false;
    }
    for (auto &p : rs)
      if (p.first < 0 || p.second < 0) good = false;
    if (!good) {
      cout << -1 << '\n';
      continue;
    }
    vector<int> rotated(n, -1);
    vector<int> queue;
    for (int i = 0; i < n; i++) {
      if (rotated[i] < 0) {
        int rev = 0;
        int cnt = 1;
        int start = b[i].first;
        int end = b[i].second;
        int j = i;
        bool fst = true;
        bool prev = false;
        rotated[i] = 0;
        queue.clear();
        queue.push_back(j);
        while (start != end) {
          if (rs[start].first == j) {
            j = rs[start].second;
          } else {
            j = rs[start].first;
          }
          if (b[j].first == start) {
            start = b[j].second;
            rotated[j] = (fst xor prev) ? 1 : 0;
            rev += rotated[j];
            fst = false;
          } else {
            start = b[j].first;
            rotated[j] = (fst xor prev) ? 0 : 1;
            rev += rotated[j];
            fst = true;
          }
          prev = rotated[j] == 1;
          cnt += 1;
          queue.push_back(j);
        }
        if (cnt - rev < rev) {
          for (auto &k : queue) rotated[k] = 1 - rotated[k];
        }
      }
    }
    if (good) {
      int res = 0;
      for (auto &x : rotated) res += x;
      cout << res << '\n';
      for (int i = 0; i < n; i++) {
        if (rotated[i] == 1) cout << i + 1 << ' ';
      }
      cout << '\n';
    } else {
      cout << -1 << '\n';
    }
  }
  return 0;
}
