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 solve() {
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  vector<vector<int>> ind(n + 1);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ind[a[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    ind[b[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (ind[i].size() != 2) {
      cout << "-1\n";
      return;
    }
  }
  vector<bool> used(n);
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    if (used[i]) continue;
    used[i] = true;
    pair<int, int> p{a[i], b[i]};
    vector<int> fi, si;
    si.push_back(i);
    while (p.first != p.second) {
      int i1 = used[ind[p.first][0]] ? ind[p.first][1] : ind[p.first][0];
      int i2 = used[ind[p.second][0]] ? ind[p.second][1] : ind[p.second][0];
      if (a[i1] == p.first) {
        p.first = b[i1];
        fi.push_back(i1);
      } else {
        p.first = a[i1];
        si.push_back(i1);
      }
      used[i1] = true;
      if (i1 == i2) break;
      if (b[i2] == p.second) {
        p.second = a[i2];
        fi.push_back(i2);
      } else {
        p.second = b[i2];
        si.push_back(i2);
      }
      used[i2] = true;
    }
    if (fi.size() < si.size())
      for (int i = 0; i < fi.size(); i++) ans.push_back(fi[i]);
    else
      for (int i = 0; i < si.size(); i++) ans.push_back(si[i]);
  }
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] + 1 << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cout << setprecision(12);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
