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 <typename T>
void setmax(T& x, T y) {
  x = max(x, y);
}
template <typename T>
void setmin(T& x, T y) {
  x = min(x, y);
}
const double PI = acos(-1.0);
const int INF = 1e9 + 47;
const long long LINF = 1LL * INF * INF;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int a[2][1 << 20];
vector<int> pos[1 << 20];
vector<int> g[1 << 20][2];
int u[1 << 20];
vector<int> comp[2];
void dfs(int v, int t = 0) {
  comp[t].push_back(v);
  u[v] = 1;
  for (int to : g[v][0])
    if (!u[to]) dfs(to, t);
  for (int to : g[v][1])
    if (!u[to]) dfs(to, 1 - t);
}
void solve() {
  int n;
  cin >> n;
  for (int i = (0); i < (n); ++i) {
    pos[i + 1].clear();
    for (int j = (0); j < (2); ++j) {
      g[i][j].clear();
    }
    u[i] = 0;
  }
  for (int i = (0); i < (2); ++i) {
    for (int j = (0); j < (n); ++j) {
      cin >> a[i][j];
      pos[a[i][j]].push_back(i * n + j);
    }
  }
  for (int i = (0); i < (n); ++i) {
    if ((int)((pos[i + 1]).size()) != 2) {
      cout << -1 << endl;
      return;
    }
    int ok1 = (pos[i + 1][0] < n);
    int ok2 = (pos[i + 1][1] < n);
    for (int j = (0); j < (2); ++j)
      g[pos[i + 1][j] % n][1 - (ok1 ^ ok2)].push_back(pos[i + 1][1 - j] % n);
  }
  vector<int> ans;
  for (int i = (0); i < (n); ++i) {
    if (u[i]) continue;
    for (int j = (0); j < (2); ++j) comp[j].clear();
    dfs(i);
    int t = ((int)((comp[1]).size()) < (int)((comp[0]).size()));
    for (int v : comp[t]) ans.push_back(v);
  }
  cout << ((int)((ans).size())) << endl;
  for (int i = (0); i < ((int)((ans).size())); ++i) {
    cout << ans[i] % n + 1 << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int q;
  cin >> q;
  while (q--) {
    solve();
  }
  cerr << "Time elapsed: " << clock() / (double)CLOCKS_PER_SEC << endl;
  return 0;
}
