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<vector<pair<int, int>>> g;
vector<int> used;
vector<int> ord1, ord2;
void dfs(int v, int col = 1) {
  used[v] = col;
  if (col > 0)
    ord1.push_back(v);
  else
    ord2.push_back(v);
  for (auto &u : g[v])
    if (!used[u.first]) dfs(u.first, col * u.second);
}
void solve() {
  int n;
  cin >> n;
  g.resize(n);
  used.resize(n);
  vector<int> a1(n), a2(n);
  vector<vector<pair<int, int>>> pos(n);
  for (int i = 0; i < n; ++i) {
    cin >> a1[i];
    a1[i]--;
    pos[a1[i]].emplace_back(i, 1);
  }
  for (int i = 0; i < n; ++i) {
    cin >> a2[i];
    a2[i]--;
    pos[a2[i]].emplace_back(i, 2);
  }
  for (int i = 0; i < n; ++i) {
    if (pos[i].size() != 2) {
      cout << "-1\n";
      used.clear();
      g.clear();
      return;
    }
    int a = pos[i][0].first, b = pos[i][1].first, c;
    if (a == b) {
      used[a] = 1;
      continue;
    }
    if (pos[i][0].second != pos[i][1].second)
      c = 1;
    else
      c = -1;
    g[a].emplace_back(b, c);
    g[b].emplace_back(a, c);
  }
  vector<int> out;
  for (int i = 0; i < n; ++i) {
    if (!used[i]) {
      ord1.clear();
      ord2.clear();
      dfs(i);
      if (ord1.size() > ord2.size()) ord1.swap(ord2);
      for (int j : ord1) out.push_back(j);
    }
  }
  cout << out.size() << '\n';
  for (int i : out) cout << i + 1 << ' ';
  cout << '\n';
  used.clear();
  g.clear();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve();
}
