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;
vector<long long> a, b, u, z, p, q;
vector<vector<long long> > g, c;
void dfs(int v, int t) {
  u[v] = 1;
  (t ? q : p).push_back(v);
  for (auto to : g[v]) {
    if (u[to]) continue;
    if (a[v] != a[to] && b[v] != b[to])
      dfs(to, t);
    else
      dfs(to, t ^ 1);
  }
}
void f() {
  cin >> n;
  a = b = vector<long long>(n);
  g = c = vector<vector<long long> >(n, vector<long long>());
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    --a[i];
    c[a[i]].push_back(i);
  }
  for (int i = 0; i < n; ++i) {
    cin >> b[i];
    --b[i];
    c[b[i]].push_back(i);
  }
  for (auto x : c) {
    if (x.size() != 2) {
      cout << -1 << "\n";
      return;
    }
  }
  for (int i = 0; i < n; ++i) {
    if (c[i][0] == c[i][1]) continue;
    int x = c[i][0];
    int y = c[i][1];
    g[x].push_back(y);
    g[y].push_back(x);
  }
  z.clear();
  u.assign(n, 0);
  for (int i = 0; i < n; ++i) {
    if (u[i] || g[i].empty()) continue;
    p.clear();
    q.clear();
    dfs(i, 0);
    if (p.size() > q.size()) swap(p, q);
    for (auto x : p) z.push_back(x);
  }
  cout << z.size() << "\n";
  for (auto x : z) cout << x + 1 << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) f();
  return 0;
}
