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;
const int N = 500005;
int n, a[3][N], used[N];
vector<int> pos[N];
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    pos[i].clear();
    used[i] = 0;
  }
  for (int i = 1; i <= n; i++) cin >> a[1][i];
  for (int i = 1; i <= n; i++) cin >> a[2][i];
  for (int i = 1; i <= n; i++) {
    pos[a[1][i]].push_back(i);
    pos[a[2][i]].push_back(i);
  }
  for (int i = 1; i <= n; i++)
    if (pos[i].size() != 2) {
      cout << -1 << endl;
      return;
    }
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    vector<int> x[2];
    int v = i, val = a[1][i];
    if (used[i]) continue;
    do {
      x[(a[1][v] != val)].push_back(v);
      used[v] = 1;
      val = a[1][v] + a[2][v] - val;
      v = (pos[val][0] + pos[val][1] - v);
    } while (!used[v]);
    if (x[1].size() < x[0].size()) swap(x[1], x[0]);
    for (int j = 0; j < x[0].size(); j++) ans.push_back(x[0][j]);
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}
