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 PII = pair<int, int>;
int const N = 3 * 2e5 + 10;
bool can_do = true;
vector<PII> g[N];
int vis[N];
vector<int> pairs[2];
void dfs(int u) {
  for (auto v : g[u]) {
    int vv = v.first;
    if (vis[vv] == -1) vis[vv] = vis[u] ^ v.second, dfs(v.first);
    can_do &= (vis[vv] ^ vis[u]) == v.second;
  }
  pairs[vis[u]].push_back(u);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    can_do = true;
    for (int i = 0; i < 3 * n; i++) g[i].clear();
    vector<vector<int>> a(2, vector<int>(n + 1));
    vector<vector<PII>> occur(n);
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < n; j++) {
        cin >> a[i][j];
        a[i][j]--;
        occur[a[i][j]].push_back({j, i});
      }
    }
    for (int i = 0; i < n; i++) {
      if (occur[i].size() != 2) can_do = false;
    }
    if (!can_do) {
      cout << "-1\n";
      continue;
    }
    for (int i = 0; i < n; i++) {
      assert(occur[i].size() == 2);
      PII u = occur[i][0], v = occur[i][1];
      if (u.first % n == v.first % n) continue;
      int tp = u.second == v.second;
      g[u.first].push_back({v.first, tp}), g[v.first].push_back({u.first, tp});
    }
    fill(vis, vis + n + 1, -1);
    vector<int> ans;
    for (int i = 0; i < n; i++) {
      if (vis[i] == -1) {
        vis[i] = 0;
        pairs[0].clear(), pairs[1].clear();
        dfs(i);
        if (pairs[0].size() > pairs[1].size()) swap(pairs[0], pairs[1]);
        while (pairs[0].size())
          ans.push_back(pairs[0].back()), pairs[0].pop_back();
      }
    }
    if (!can_do) {
      cout << "-1\n";
      continue;
    }
    cout << ans.size() << '\n';
    for (int x : ans) cout << x + 1 << " ";
    cout << '\n';
  }
}
