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<int> Not[400005];
vector<int> where[400005];
vector<int> tt[5];
vector<int> ans;
int color[400005];
bool have[400005] = {0};
int con[200005];
int all[5][200005];
int N, ok, a, b;
void F(int here, int co) {
  if (have[here]) {
    if (color[here] != co) ok = 0;
  } else {
    if (here < N) {
      if (co == 0) {
        tt[0].push_back(here);
        a++;
      } else {
        tt[1].push_back(here);
        b++;
      }
    }
    have[here] = 1;
    color[here] = co;
    for (auto i : Not[here]) F(i, 1 - co);
  }
}
int main() {
  int T, i, j, how = 0;
  scanf("%d", &T);
  while (T--) {
    how++;
    ans.clear();
    ok = 1;
    a = b = 0;
    scanf("%d", &N);
    for (i = 1; i <= N; i++) con[i] = 0;
    for (i = 0; i < 2 * N; i++) {
      Not[i].clear();
      where[i].clear();
    }
    for (i = 0; i < N; i++) {
      Not[i].push_back(i + N);
      Not[i + N].push_back(i);
    }
    for (i = 0; i < N; i++) {
      scanf("%d", &all[0][i]);
      con[all[0][i]]++;
      where[all[0][i]].push_back(i);
      have[i] = 0;
    }
    for (i = 0; i < N; i++) {
      scanf("%d", &all[1][i]);
      con[all[1][i]]++;
      where[all[1][i]].push_back(i + N);
      have[i + N] = 0;
    }
    for (i = 1; i <= N; i++)
      if (con[i] != 2) ok = 0;
    if (ok) {
      for (i = 1; i <= N; i++) {
        Not[where[i][0]].push_back(where[i][1]);
        Not[where[i][1]].push_back(where[i][0]);
      }
      for (i = 0; i < 2 * N; i++) {
        if (have[i]) continue;
        tt[0].clear();
        tt[1].clear();
        a = b = 0;
        F(i, 0);
        if (a < b)
          for (auto i : tt[0]) ans.push_back(i);
        else
          for (auto i : tt[1]) ans.push_back(i);
      }
      if (ok) {
        printf("%d\n", ans.size());
        for (auto i : ans) printf("%d ", i + 1);
        printf("\n");
      } else
        printf("-1\n");
    } else
      printf("-1\n");
  }
  return 0;
}
