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> l, r, vis, nf, f;
vector<vector<int>> g;
vector<pair<int, int>> p;
void solve(int x) {
  for (auto v : g[x]) {
    if (vis[v]) continue;
    vis[v] = 1;
    if (l[p[v].first] || r[p[v].second]) {
      swap(p[v].first, p[v].second);
      f.push_back(v);
    } else
      nf.push_back(v);
    l[p[v].first] = 1;
    r[p[v].second] = 1;
    solve(p[v].first);
    solve(p[v].second);
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    f.clear();
    nf.clear();
    p.resize(n);
    l.assign(n + 1, 0);
    r.assign(n + 1, 0);
    vis.assign(n, 0);
    g.assign(n + 1, vector<int>());
    for (int i = 0; i < n; ++i) {
      cin >> p[i].first;
      g[p[i].first].push_back(i);
    }
    for (int i = 0; i < n; ++i) {
      cin >> p[i].second;
      g[p[i].second].push_back(i);
    }
    bool poss = 1;
    for (int i = 1; i <= n; ++i) {
      poss &= ((int)g[i].size() == 2);
    }
    if (!poss) {
      cout << -1 << '\n';
      continue;
    }
    vector<int> ans;
    for (int i = 0; i < n; ++i) {
      if (vis[i]) continue;
      if (f.size() > nf.size()) swap(f, nf);
      for (auto i : f) ans.push_back(i);
      f.clear();
      nf.clear();
      nf.push_back(i);
      vis[i] = 1;
      l[p[i].first] = 1;
      r[p[i].second] = 1;
      solve(p[i].first);
      solve(p[i].second);
    }
    if (f.size() > nf.size()) swap(f, nf);
    for (auto i : f) ans.push_back(i);
    cout << ans.size() << '\n';
    for (auto i : ans) cout << i + 1 << " ";
    cout << '\n';
  }
}
