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 std::cin;
using std::cout;
using std::endl;
using std::string;
using std::vector;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int flt = 0;
    vector<int> a, b;
    vector<int> pos1, pos2;
    for (int i = 0; i < n; ++i) {
      int c;
      cin >> c;
      a.push_back(c);
    }
    for (int i = 0; i < n; ++i) {
      int c;
      cin >> c;
      b.push_back(c);
      pos1.push_back(-1);
      pos2.push_back(-1);
    }
    for (int i = 0; i < n; ++i) {
      if (pos1[a[i] - 1] == -1) {
        pos1[a[i] - 1] = i;
      } else {
        if (pos2[a[i] - 1] == -1) {
          pos2[a[i] - 1] = i;
        } else {
          flt = 1;
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      if (pos1[b[i] - 1] == -1) {
        pos1[b[i] - 1] = i;
      } else {
        if (pos2[b[i] - 1] == -1) {
          pos2[b[i] - 1] = i;
        } else {
          flt = 1;
        }
      }
    }
    if (flt == 1) {
      cout << "-1" << endl;
    } else {
      vector<int> ans;
      vector<int> izok;
      for (int i = 0; i < n; ++i) {
        izok.push_back(0);
      }
      for (int i = 0; i < n; ++i) {
        if (izok[i] == 0) {
          vector<int> ans1, ans2;
          int nextpos = -1;
          int j = i;
          ans2.push_back(i);
          while (nextpos != i) {
            nextpos = (pos1[b[j] - 1] == j) ? pos2[b[j] - 1] : pos1[b[j] - 1];
            if (nextpos == i) {
              izok[i] = 1;
            } else {
              if ((a[j] == a[nextpos]) || (b[j] == b[nextpos])) {
                ans1.push_back(nextpos);
                int temp = b[nextpos];
                b[nextpos] = a[nextpos];
                a[nextpos] = temp;
                izok[nextpos] = 1;
                j = nextpos;
              } else {
                ans2.push_back(nextpos);
                izok[nextpos] = 1;
                j = nextpos;
              }
            }
          }
          if (ans1.size() < ans2.size()) {
            for (int l = 0; l < ans1.size(); ++l) {
              ans.push_back(ans1[l]);
            }
          } else {
            for (int l = 0; l < ans2.size(); ++l) {
              ans.push_back(ans2[l]);
            }
          }
          ans1.clear();
          ans2.clear();
        }
      }
      cout << ans.size() << endl;
      for (int i = 0; i < ans.size(); ++i) {
        cout << ans[i] + 1 << " ";
      }
      cout << endl;
      ans.clear();
      izok.clear();
    }
    a.clear();
    b.clear();
    pos1.clear();
    pos2.clear();
  }
  return 0;
}
