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;
vector<int> c;
vector<vector<int>> z;
vector<vector<vector<int>>> edges;
vector<int> a, b;
void dfs(int ver) {
  for (auto e : edges[ver]) {
    if (!c[e[0]]) {
      if (e[1]) {
        if (c[ver] == 1)
          c[e[0]] = 2, b.push_back(e[0]);
        else
          c[e[0]] = 1, a.push_back(e[0]);
      } else {
        if (c[ver] == 1)
          c[e[0]] = 1, a.push_back(e[0]);
        else
          c[e[0]] = 2, b.push_back(e[0]);
      }
      dfs(e[0]);
    }
  }
}
inline void test_case() {
  int n;
  cin >> n;
  z.assign(n + 1, {});
  for (int r = 0; r < 2; r++) {
    for (int c = 1; c <= n; c++) {
      int x;
      cin >> x;
      z[x].push_back(r);
      z[x].push_back(c);
    }
  }
  edges.assign(n + 1, {});
  for (int i = 1; i <= n; i++) {
    if (z[i].size() != 4) {
      cout << "-1\n";
      return;
    }
    edges[z[i][1]].push_back({z[i][3], (z[i][0] == z[i][2] ? 1 : 0)});
    edges[z[i][3]].push_back({z[i][1], (z[i][0] == z[i][2] ? 1 : 0)});
  }
  vector<int> ans;
  c.assign(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    if (!c[i]) {
      c[i] = 1, a.push_back(i);
      dfs(i);
      if (a.size() < b.size()) {
        for (int j : a) ans.push_back(j);
      } else {
        for (int j : b) ans.push_back(j);
      }
      a.clear(), b.clear();
    }
  }
  cout << ans.size() << '\n';
  for (int i : ans) cout << i << " ";
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  for (int i = 1; i <= t; i++) test_case();
  return 0;
}
