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>
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("popcnt,abm,mmx,avx2")
using namespace std;
template <class U, class V>
istream& operator>>(istream& is, pair<U, V>& p) {
  is >> p.first >> p.second;
  return is;
}
template <class T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& x : v) {
    is >> x;
  }
  return is;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto& x : v) {
    os << x << " ";
  }
  return os;
}
template <typename T, typename... Ts>
T min(const T& h1, const T& h2, const Ts&... tail) {
  return min(min(h1, h2), tail...);
}
template <typename T, typename... Ts>
T max(const T& h1, const T& h2, const Ts&... tail) {
  return max(max(h1, h2), tail...);
}
const int N = 2e5 + 5;
vector<pair<int, bool>> adj[N];
int n;
int color[N];
vector<int> taken, not_taken;
int dfs(int node, bool c) {
  if (color[node] == c) return 0;
  if (color[node] == !c) return N;
  if (c) {
    taken.emplace_back(node);
  } else {
    not_taken.emplace_back(node);
  }
  int res = c;
  color[node] = c;
  for (auto [v, x] : adj[node]) {
    res += dfs(v, c ^ x);
  }
  return min(res, N);
}
void init() {
  for (int i = 0; i < n; ++i) {
    adj[i].clear();
    color[i] = -1;
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<int> a(n), b(n);
    init();
    cin >> a >> b;
    vector<vector<pair<int, int>>> occ(n);
    for (int i = 0; i < n; ++i) {
      occ[a[i] - 1].emplace_back(0, i);
      occ[b[i] - 1].emplace_back(1, i);
    }
    bool valid = true;
    for (int i = 0; i < n; ++i) {
      valid &= (((int)(occ[i].size())) == 2);
    }
    if (!valid) {
      cout << -1 << '\n';
      continue;
    }
    for (int i = 0; i < n; ++i) {
      if (occ[i][0].first == occ[i][1].first) {
        adj[occ[i][0].second].emplace_back(occ[i][1].second, true);
        adj[occ[i][1].second].emplace_back(occ[i][0].second, true);
      } else {
        adj[occ[i][0].second].emplace_back(occ[i][1].second, false);
        adj[occ[i][1].second].emplace_back(occ[i][0].second, false);
      }
    }
    vector<int> res;
    for (int i = 0; i < n; ++i) {
      if (color[i] != -1) continue;
      taken.clear();
      not_taken.clear();
      int tmp = dfs(i, true);
      if (tmp == N) {
        res = {-1};
        break;
      }
      if (((int)(taken.size())) < ((int)(not_taken.size()))) {
        res.insert(end(res), begin(taken), end(taken));
      } else {
        res.insert(end(res), begin(not_taken), end(not_taken));
      }
    }
    if (res == vector<int>{-1}) {
      cout << -1 << '\n';
    } else {
      cout << ((int)(res.size())) << '\n';
      for (int x : res) {
        cout << x + 1 << " ";
      }
      cout << '\n';
    }
  }
  return 0;
}
