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;
vector<long long> pos[200005], cost[400005];
long long arr[2][200005], p[400005], n, vis[400005];
inline long long no(long long x) {
  if (x <= n)
    return x + n;
  else
    return x - n;
}
void init(long long n) {
  for (long long i = 1; i <= 2 * n; i++) p[i] = i, vis[i] = 0, cost[i].clear();
}
long long fp(long long x) {
  if (x == p[x]) return x;
  return p[x] = fp(p[x]);
}
void merge(long long x, long long y) {
  x = fp(x), y = fp(y);
  if (x == y) return;
  p[x] = y;
}
void solve() {
  cin >> n;
  init(n);
  for (long long i = 1; i <= n; i++) pos[i].clear();
  for (long long i = 1; i <= n; i++) {
    cin >> arr[0][i];
    pos[arr[0][i]].push_back(i);
  }
  for (long long i = 1; i <= n; i++) {
    cin >> arr[1][i];
    pos[arr[1][i]].push_back(i + n);
  }
  for (long long i = 1; i <= n; i++) {
    if (pos[i].size() != 2) {
      cout << -1 << '\n';
      return;
    }
  }
  for (long long i = 1; i <= n; i++) {
    long long x = pos[i][0], y = pos[i][1];
    merge(x, no(y));
    merge(no(x), y);
  }
  for (long long i = 1; i <= n; i++) {
  }
  for (long long i = n + 1; i <= 2 * n; i++) {
    cost[fp(i)].push_back(no(i));
  }
  long long ans = 0;
  vector<long long> aa;
  for (long long i = 1; i <= n; i++) {
    if (vis[fp(i)]) continue;
    vis[fp(i)] = vis[fp(no(i))] = 1;
    if (cost[fp(i)].size() < cost[fp(no(i))].size()) {
      ans += cost[fp(i)].size();
      for (auto x : cost[fp(i)]) aa.push_back(x);
    } else {
      ans += cost[fp(no(i))].size();
      for (auto x : cost[fp(no(i))]) aa.push_back(x);
    }
  }
  cout << ans << '\n';
  for (auto i : aa) {
    cout << i << " ";
  }
  cout << '\n';
}
signed main() {
  ios::sync_with_stdio(0), cin.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) solve();
}
