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 maxn = 2e5 + 5;
int N, a[2][maxn];
vector<pair<int, int> > pos[maxn], adj[maxn];
void solve(void) {
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    pos[i].clear();
    adj[i].clear();
  }
  for (int i = 0; i < 2; ++i) {
    for (int j = 1; j <= N; ++j) {
      cin >> a[i][j];
      pos[a[i][j]].emplace_back(i, j);
    }
  }
  for (int i = 1; i <= N; ++i) {
    if (pos[i].size() != 2) {
      cout << "-1\n";
      return;
    }
    pair<int, int> p1 = pos[i][0], p2 = pos[i][1];
    if (p1.second == p2.second) continue;
    adj[p1.second].emplace_back(p2.second, (p1.first == p2.first));
    adj[p2.second].emplace_back(p1.second, (p1.first == p2.first));
  }
  vector<int> col(N + 5, -1);
  vector<int> same[2], res;
  function<void(int)> dfs = [&](int u) {
    same[col[u]].emplace_back(u);
    for (auto& v : adj[u]) {
      if (col[v.first] == -1) {
        col[v.first] = col[u] ^ v.second;
        dfs(v.first);
      }
    }
  };
  for (int i = 1; i <= N; ++i) {
    if (col[i] == -1) {
      same[0].clear();
      same[1].clear();
      col[i] = 0;
      dfs(i);
      if (same[0].size() > same[1].size()) swap(same[0], same[1]);
      for (auto& j : same[0]) res.emplace_back(j);
    }
  }
  cout << res.size() << '\n';
  for (auto& i : res) cout << i << ' ';
  cout << '\n';
}
signed main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    solve();
  }
}
