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;
int A[200003];
int B[200003];
int col[200003];
bool visited[200003];
vector<int> ans;
vector<int> cols[2];
vector<int> vec[200003];
vector<int> pos[200003];
bool beda(int u, int v) { return A[u] == A[v] || B[u] == B[v]; }
bool fail = 0;
void dfs(int u, int c) {
  col[u] = c;
  cols[c].push_back(u);
  visited[u] = 1;
  for (int v : vec[u]) {
    if (visited[v]) {
      if (col[u] ^ beda(u, v) != col[v]) fail = 1;
      continue;
    }
    dfs(v, c ^ beda(u, v));
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  scanf("%d", &t);
  while (t--) {
    fail = 0;
    ans.clear();
    for (int i = 1; i <= n; i++) {
      col[i] = 0;
      visited[i] = 0;
      pos[i].clear();
      vec[i].clear();
    }
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      int x;
      scanf("%d", &x);
      A[i] = x;
      pos[x].push_back(i);
    }
    for (int i = 1; i <= n; i++) {
      int x;
      scanf("%d", &x);
      B[i] = x;
      pos[x].push_back(i);
    }
    bool bol = 1;
    for (int i = 1; i <= n; i++) {
      if (pos[i].size() != 2) {
        bol = 0;
        break;
      }
      if (pos[i][0] == pos[i][1]) continue;
      vec[pos[i][0]].push_back(pos[i][1]);
      vec[pos[i][1]].push_back(pos[i][0]);
    }
    if (!bol) {
      printf("-1\n");
      continue;
    }
    int res = 0;
    for (int i = 1; i <= n; i++) {
      if (!visited[i]) {
        cols[0].clear();
        cols[1].clear();
        dfs(i, 0);
        if (cols[0].size() < cols[1].size())
          for (auto x : cols[0]) ans.push_back(x);
        else
          for (auto x : cols[1]) ans.push_back(x);
      }
    }
    if (fail) {
      printf("-1\n");
      continue;
    }
    printf("%d\n", ans.size());
    for (int i = 0; i < ans.size(); i++) {
      printf("%d%c", ans[i], (i == ans.size() - 1) ? '\n' : ' ');
    }
    if (ans.size() == 0) printf("\n");
  }
}
