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("O3", "unroll-loops")
using namespace std;
using pii = pair<long long, long long>;
template <typename T>
using prior = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using Prior = priority_queue<T>;
const long long INF = 1E18;
const long long mod = 1E9 + 7;
void solve() {
  long long n;
  cin >> n;
  vector<long long> v1(n), v2(n), cnt(n + 1);
  for (auto &x : v1) cin >> x;
  for (auto &x : v2) cin >> x;
  vector<pii> adj[n + 1];
  for (long long i = 0; i < n; ++i) {
    if (++cnt[v1[i]] > 2 or ++cnt[v2[i]] > 2) {
      cout << -1 << "\n";
      return;
    }
    adj[v1[i]].emplace_back(v2[i], i + 1),
        adj[v2[i]].emplace_back(-v1[i], i + 1);
  }
  vector<long long> vis(n + 1), ans;
  for (long long i = 1; i <= n; ++i) {
    long long now = i, lst = -1;
    vector<long long> pos, neg;
    while (!vis[now]) {
      vis[now] = 1;
      if (abs(adj[now][0].first) != lst) {
        if (adj[now][0].first > 0)
          pos.emplace_back(adj[now][0].second);
        else
          neg.emplace_back(adj[now][0].second);
        lst = now, now = abs(adj[now][0].first);
      } else {
        if (adj[now][1].first > 0)
          pos.emplace_back(adj[now][1].second);
        else
          neg.emplace_back(adj[now][1].second);
        lst = now, now = abs(adj[now][1].first);
      }
    }
    if (pos.size() <= neg.size()) {
      for (auto x : pos) ans.emplace_back(x);
    } else {
      for (auto x : neg) ans.emplace_back(x);
    }
  }
  cout << ans.size() << "\n";
  for (auto x : ans) cout << x << " ";
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}
