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> g[200100], d[200100];
bool vis[200100];
int a[2][200100];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < n; j++) {
        scanf("%d", &a[i][j]);
      }
    }
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < n; j++) {
        g[a[i][j]].push_back(a[i ^ 1][j]);
        d[a[i][j]].push_back(j << 1 | i);
      }
    }
    bool fail = false;
    for (int i = 1; i <= n; i++) {
      if (g[i].size() != 2) fail = true;
    }
    if (fail)
      puts("-1");
    else {
      vector<int> ans;
      for (int u = 1; u <= n; u++) {
        vector<int> v[2];
        int p = -1;
        while (!vis[u]) {
          vis[u] = true;
          if (d[u][0] == (p ^ 1)) {
            p = d[u][1];
            u = g[u][1];
          } else {
            p = d[u][0];
            u = g[u][0];
          }
          v[p & 1].push_back(p >> 1);
        }
        if (v[0].size() < v[1].size())
          ans.insert(ans.end(), v[0].begin(), v[0].end());
        else
          ans.insert(ans.end(), v[1].begin(), v[1].end());
      }
      printf("%d\n", (int)ans.size());
      sort(ans.begin(), ans.end());
      for (auto x : ans) printf("%d ", x + 1);
      puts("");
    }
    for (int i = 1; i <= n; i++) {
      g[i].clear();
      d[i].clear();
      vis[i] = false;
    }
  }
}
