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;
template <class T1, class T2>
ostream& operator<<(ostream& out, const pair<T1, T2> p) {
  out << '(' << p.first << ',' << p.second << ')';
  return out;
}
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
template <class T>
istream& operator>>(istream& in, vector<T>& v) {
  for (T& x : v) in >> x;
  return in;
}
template <class T>
ostream& operator<<(ostream& out, const vector<vector<T>>& v) {
  for (const vector<T>& x : v) out << x << '\n';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  for (const T& x : v) out << x << ' ';
  return out;
}
long long gcd(long long a, long long b) {
  if (b > a) swap(a, b);
  return (b ? gcd(b, a % b) : a);
}
using ll = long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using tiii = pair<pair<int, int>, int>;
using vi = vector<int>;
using vl = vector<long long>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<long long>>;
const int h = 1000000007;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.setstate(ios::failbit);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vi v(n), v2(n), cnt(n);
    cin >> v >> v2;
    for (int x : v) cnt[x - 1]++;
    for (int x : v2) cnt[x - 1]++;
    bool y = 1;
    for (int x : cnt)
      if (x != 2) y = 0;
    if (!y) {
      cout << "-1\n";
      continue;
    }
    vi vis(n);
    vvi ind(n);
    for (int i = (0); i < (n); i++) ind[v[i] - 1].push_back(i);
    for (int i = (0); i < (n); i++) ind[v2[i] - 1].push_back(i);
    int ans = 0;
    vi final_inds;
    for (int i = (0); i < (n); i++) {
      int x = i;
      if (vis[x]) continue;
      int val = v[x] - 1;
      set<int> all_inds, swapped_inds;
      while (!vis[x]) {
        all_inds.insert(x + 1);
        vis[x] = 1;
        x = ind[val][ind[val][0] == x];
        if (v[x] != val + 1) {
          swapped_inds.insert(x + 1);
          swap(v[x], v2[x]);
        }
        val = v2[x] - 1;
      }
      if (swapped_inds.size() < all_inds.size() - swapped_inds.size()) {
        for (int x : swapped_inds) final_inds.push_back(x);
        ans += swapped_inds.size();
      } else {
        for (int x : all_inds)
          if (!swapped_inds.count(x)) final_inds.push_back(x);
        ans += all_inds.size() - swapped_inds.size();
      }
    }
    cout << ans << '\n' << final_inds << '\n';
  }
}
