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;
using ill = long long;
class solution {
  vector<int> col, pth;
  vector<vector<pair<int, int>>> g;
  bool dfs(int u, int las) {
    pth.push_back(u);
    for (auto v : g[u]) {
      if (v.first == las) continue;
      int nxtCol = -1;
      if (v.second == 1)
        nxtCol = col[u];
      else
        nxtCol = 1 - col[u];
      if (col[v.first] != -1) {
        return col[v.first] == nxtCol;
      }
      col[v.first] = nxtCol;
      if (dfs(v.first, u) == false) return false;
    }
    return true;
  }

 public:
  void solve() {
    int tct;
    cin >> tct;
    for (int tcc = 1; tcc <= tct; ++tcc) {
      int n;
      cin >> n;
      if (tcc == 17) {
        for (int __s = 0; __s == 0; ++__s)
          ;
      }
      vector<int> cntX(n + 1, 0);
      vector<vector<pair<int, int>>> pos(n + 1);
      for (int i = 0; i <= n - 1; ++i) {
        int tmp;
        cin >> tmp;
        ++cntX[tmp];
        pos[tmp].push_back({i, 0});
      }
      for (int i = 0; i <= n - 1; ++i) {
        int tmp;
        cin >> tmp;
        ++cntX[tmp];
        pos[tmp].push_back({i, 1});
      }
      bool flg = true;
      for (int x = 1; x <= n; ++x) {
        if (cntX[x] != 2) {
          flg = false;
        }
      }
      if (flg == false) {
        cout << "-1\n";
        continue;
      }
      col.assign(n, -1);
      g.assign(n, vector<pair<int, int>>());
      for (int x = 1; x <= n; ++x) {
        g[pos[x][0].first].push_back(
            {pos[x][1].first, pos[x][0].second != pos[x][1].second});
        g[pos[x][1].first].push_back(
            {pos[x][0].first, pos[x][0].second != pos[x][1].second});
      }
      vector<int> ans;
      for (int u = 0; u <= n - 1; ++u) {
        if (col[u] == -1) {
          col[u] = 0;
          pth.clear();
          if (dfs(u, -1) == false) {
            flg = false;
            break;
          }
          vector<int> tmp;
          for (auto v : pth)
            if (col[v] == 1) tmp.push_back(v + 1);
          if (tmp.size() > pth.size() - tmp.size()) {
            tmp.clear();
            for (auto v : pth)
              if (col[v] == 0) tmp.push_back(v + 1);
          }
          for (auto v : tmp) ans.push_back(v);
        }
      }
      if (flg == true) {
        int siz = ans.size();
        cout << siz << '\n';
        for (int i = 0; i <= siz - 1; ++i) {
          if (i > 0) cout << ' ';
          cout << ans[i];
        }
        cout << '\n';
      } else {
        cout << "-1\n";
      }
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  solution sln1;
  sln1.solve();
  cout.flush();
  return 0;
}
