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;
int N, root[200005], A[200005], B[200005];
vector<int> comp[200005], edges[200005];
vector<pair<int, bool>> pos[200005];
bool vis[200005];
int findR(int x) { return x == root[x] ? x : root[x] = findR(root[x]); }
void merge(int a, int b) {
  a = findR(a);
  b = findR(b);
  if (a == b) return;
  if (comp[a].size() < comp[b].size()) swap(a, b);
  root[b] = a;
  for (int x : comp[b]) comp[a].push_back(x);
  return;
}
void dfs(int v, bool t, int& odd, int& even, vector<int>& o, vector<int>& e) {
  if (vis[v]) return;
  vis[v] = true;
  (t ? odd : even) += comp[v].size();
  (t ? o : e).push_back(v);
  for (auto u : edges[v]) {
    dfs(u, !t, odd, even, o, e);
  }
  return;
}
void solve() {
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    root[i] = i;
    comp[i] = {i};
    pos[i].clear();
    edges[i].clear();
    vis[i] = false;
  }
  for (int i = 1; i <= N; ++i) {
    cin >> A[i];
    pos[A[i]].push_back({i, 0});
  }
  for (int i = 1; i <= N; ++i) {
    cin >> B[i];
    pos[B[i]].push_back({i, 1});
  }
  for (int i = 1; i <= N; ++i) {
    if (pos[i].size() != 2) {
      cout << -1 << endl;
      return;
    }
    if (pos[i][0].second != pos[i][1].second) {
      merge(pos[i][0].first, pos[i][1].first);
    }
  }
  for (int i = 1; i <= N; ++i) {
    if (pos[i][0].second == pos[i][1].second) {
      edges[findR(pos[i][0].first)].push_back(findR(pos[i][1].first));
      edges[findR(pos[i][1].first)].push_back(findR(pos[i][0].first));
    }
  }
  long long ans = 0;
  vector<int> vans;
  for (int i = 1; i <= N; ++i) {
    if (i != findR(i) || vis[i]) continue;
    int odd = 0;
    int even = 0;
    vector<int> e, o;
    dfs(i, 0, odd, even, o, e);
    if (odd < even) {
      for (int x : o) vans.push_back(x);
    } else {
      for (int x : e) vans.push_back(x);
    }
    ans += min(odd, even);
  }
  cout << ans << "\n";
  for (int c : vans) {
    for (int v : comp[c]) {
      cout << v << " ";
    }
  }
  cout << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) solve();
}
