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;
long long T, N, a[3][200005], c, M;
bool v[200005];
vector<int> node[200005], sp, nsp, ans;
void init() {
  memset(v, 0, sizeof(v));
  for (int i = 1; i <= N; ++i) node[i].clear();
  ans.clear();
}
bool check() {
  int ct[200005] = {0};
  for (int i = 1; i <= 2; ++i)
    for (int j = 1; j <= N; ++j) ct[a[i][j]]++;
  for (int i = 1; i <= N; ++i)
    if (ct[i] != 2) return false;
  return true;
}
void f(int x) {
  v[x] = 1, ++M;
  for (int i = 1; i <= 2; ++i)
    for (auto j : node[a[i][x]]) {
      if (v[j]) continue;
      if (a[i][j] == a[i][x]) {
        swap(a[1][j], a[2][j]), ++c;
        sp.emplace_back(j);
      } else
        nsp.emplace_back(j);
      f(j);
    }
}
void start(int x) {
  c = M = 0;
  nsp.clear(), sp.clear();
  nsp.emplace_back(x);
  f(x);
  if (c < M - c)
    for (auto i : sp) ans.emplace_back(i);
  else
    for (auto i : nsp) ans.emplace_back(i);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> T;
  for (int tc = 1; tc <= T; ++tc) {
    cin >> N;
    init();
    for (int i = 1; i <= 2; ++i)
      for (int j = 1; j <= N; ++j) {
        cin >> a[i][j];
        node[a[i][j]].emplace_back(j);
      }
    if (!check()) {
      cout << "-1\n";
      continue;
    }
    for (int i = 1; i <= N; ++i)
      if (!v[i]) start(i);
    cout << ans.size() << "\n";
    for (auto i : ans) cout << i << " ";
    cout << "\n";
  }
}
