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;
void dfs(vector<vector<int>>& G, vector<char>& used, vector<int>& color, int v,
         vector<int>& ans1, vector<int>& ans2) {
  used[v] = true;
  color[v] = 1;
  int other;
  if (v < G.size() / 2)
    other = v + G.size() / 2;
  else
    other = v - G.size() / 2;
  used[other] = true;
  color[other] = -1;
  if (v < other) {
    ans1.push_back(v);
  } else {
    ans2.push_back(other);
  }
  for (int u : G[v]) {
    if (!used[u]) {
      dfs(G, used, color, u, ans1, ans2);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<vector<int>> V(2, vector<int>(n));
    for (int i = 0; i < n; ++i) {
      cin >> V[0][i];
      --V[0][i];
    }
    for (int i = 0; i < n; ++i) {
      cin >> V[1][i];
      --V[1][i];
    }
    vector<int> cnt(n);
    for (int i = 0; i < n; ++i) {
      cnt[V[0][i]]++;
      cnt[V[1][i]]++;
    }
    bool ok = true;
    for (int i = 0; i < n; ++i) {
      if (cnt[i] != 2) {
        ok = false;
        break;
      }
    }
    if (!ok) {
      cout << -1 << "\n";
      continue;
    }
    vector<int> same(n, 1);
    vector<vector<int>> inds(n);
    for (int i = 0; i < n; ++i) {
      inds[V[0][i]].push_back(i);
      same[V[0][i]] ^= 1;
    }
    for (int i = 0; i < n; ++i) {
      inds[V[1][i]].push_back(i);
    }
    vector<vector<int>> G(n * 2);
    auto add_edge = [&G](int a, int b) {
      if (a != b) {
        G[a].push_back(b);
        G[b].push_back(a);
      }
    };
    for (int i = 0; i < n; ++i) {
      int a = inds[i][0];
      int b = inds[i][1];
      int na = a + n;
      int nb = b + n;
      if (same[i]) {
        add_edge(a, nb);
        add_edge(b, na);
      } else {
        add_edge(a, b);
        add_edge(na, nb);
      }
    }
    vector<int> color(n * 2, 0);
    vector<char> used(n * 2, 0);
    vector<int> ans1, ans2;
    vector<int> ans;
    for (int i = 0; i < n * 2; ++i) {
      if (!used[i]) dfs(G, used, color, i, ans1, ans2);
      if (ans1.size() < ans2.size()) {
        for (int i : ans1) ans.push_back(i);
      } else {
        for (int i : ans2) ans.push_back(i);
      }
      ans1.clear();
      ans2.clear();
    }
    cout << ans.size() << "\n";
    for (int i : ans) cout << i + 1 << ' ';
    cout << "\n";
  }
  return 0;
}
