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>
const int MAXN = 200005;
using namespace std;
int n, a[2][MAXN], col[MAXN];
vector<pair<int, int> > occ[MAXN];
vector<int> G1[MAXN], G2[MAXN], cc;
pair<int, int> dfs(int v, int c) {
  col[v] = c;
  cc.push_back(v);
  pair<int, int> my = pair<int, int>(c == 0, c == 1);
  for (int to : G1[v]) {
    if (col[to] == -1) {
      pair<int, int> res = dfs(to, c);
      if (res.first == -1) {
        return pair<int, int>(-1, -1);
      }
      my.first += res.first;
      my.second += res.second;
    }
  }
  for (int to : G2[v]) {
    if (col[to] == -1) {
      pair<int, int> res = dfs(to, !c);
      if (res.first == -1) {
        return pair<int, int>(-1, -1);
      }
      my.first += res.first;
      my.second += res.second;
    } else if (col[to] == c) {
      return pair<int, int>(-1, -1);
    }
  }
  return my;
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    col[i] = -1;
    occ[i].clear();
    G1[i].clear();
    G2[i].clear();
  }
  for (int id = 0; id <= 1; id++) {
    for (int i = 1; i <= n; i++) {
      cin >> a[id][i];
      occ[a[id][i]].push_back({id, i});
    }
  }
  for (int i = 1; i <= n; i++) {
    if (occ[i].size() != 2) {
      cout << "-1\n";
      return;
    }
    pair<int, int> p1 = occ[i][0], p2 = occ[i][1];
    if (p1.second == p2.second) {
      continue;
    }
    if (p1.first == p2.first) {
      G2[p1.second].push_back(p2.second);
      G2[p2.second].push_back(p1.second);
    } else {
      G1[p1.second].push_back(p2.second);
      G1[p2.second].push_back(p1.second);
    }
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    if (col[i] == -1) {
      cc.clear();
      pair<int, int> p = dfs(i, 0);
      if (p.first == -1) {
        cout << "-1\n";
        return;
      }
      int take = (p.second < p.first);
      for (int x : cc) {
        if (col[x] == take) {
          ans.push_back(x);
        }
      }
    }
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << " \n"[i + 1 == ans.size()];
  }
  if (ans.empty()) cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  cin >> tc;
  for (int i = 1; i <= tc; i++) {
    solve();
  }
  return 0;
}
