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;
using i64 = long long;
using u32 = unsigned int;
const int mod = 998244353;
int msum(int a, int b) { return (a + b < mod ? a + b : a + b - mod); }
int madd(int& a, int b) { a = msum(a, b); }
struct Solver {
  void Solve(istream& cin, ostream& cout) {
    int n;
    cin >> n;
    vector<int> w1(n, -1), w2(n, -1);
    vector<vector<int>> a(n, vector<int>(2));
    for (int i = 0; i < n + n; ++i) {
      int t;
      cin >> t;
      --t;
      w1[t] = w2[t];
      w2[t] = i % n;
      a[i % n][i >= n] = t;
    }
    if (find((w1).begin(), (w1).end(), -1) != w1.end() ||
        find((w2).begin(), (w2).end(), -1) != w2.end()) {
      cout << -1 << '\n';
      return;
    }
    vector<int> res;
    vector<int> notInv, inv;
    res.reserve(n);
    notInv.reserve(n);
    inv.reserve(n);
    vector<bool> was(n);
    for (int i = 0; i < n; ++i) {
      if (!was[i]) {
        was[i] = true;
        notInv.clear();
        inv.clear();
        notInv.push_back(i);
        for (int s = a[i][0], t = a[i][1], p = i; s != t;) {
          p = (w1[s] == p ? w2[s] : w1[s]);
          was[p] = true;
          if (a[p][1] == s) {
            notInv.push_back(p);
            s = a[p][0];
          } else {
            inv.push_back(p);
            s = a[p][1];
          }
        }
        if (inv.size() > notInv.size()) {
          swap(inv, notInv);
        }
        for (int x : inv) {
          res.push_back(x);
        }
      }
    }
    cout << res.size() << '\n';
    for (int x : res) {
      cout << x + 1 << ' ';
    }
    cout << '\n';
  }
};
int main() {
  int T = 1;
  cin >> T;
  for (int iTest = 1; iTest <= T; ++iTest) {
    Solver solver;
    solver.Solve(cin, cout);
  }
  return 0;
}
