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;
const int N = 2e6 + 20;
int n, x, a[N], b[N], cnt[N];
bool vis[N];
vector<int> ans, d[3];
vector<pair<int, int>> gr[N], c[N];
void dfs(int v, int col = 0) {
  d[col].push_back(v);
  vis[v] = true;
  for (auto x : gr[v]) {
    if (!vis[x.first]) dfs(x.first, (col ^ x.second));
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    ans.clear();
    for (int i = 0; i <= n; i++)
      vis[i] = false, cnt[i] = 0, c[i].clear(), gr[i].clear();
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      cnt[a[i]]++;
      c[a[i]].push_back({i, 0});
    }
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      cnt[b[i]]++;
      c[b[i]].push_back({i, 1});
    }
    bool e = true;
    for (int i = 1; i <= n; i++)
      if (cnt[i] != 2) e = false;
    if (!e)
      cout << -1 << endl;
    else {
      for (int i = 1; i <= n; i++) {
        if (c[i][0].first != c[i][1].first) {
          if (c[i][0].second == c[i][1].second)
            gr[c[i][0].first].push_back({c[i][1].first, 1}),
                gr[c[i][1].first].push_back({c[i][0].first, 1});
          else
            gr[c[i][0].first].push_back({c[i][1].first, 0}),
                gr[c[i][1].first].push_back({c[i][0].first, 0});
        }
      }
      for (int i = 0; i < n; i++) {
        if (!vis[i]) {
          d[0].clear(), d[1].clear();
          dfs(i);
          if (d[0].size() > d[1].size())
            for (auto x : d[1]) ans.push_back(x);
          else
            for (auto x : d[0]) ans.push_back(x);
        }
      }
      cout << ans.size() << endl;
      for (auto x : ans) cout << x + 1 << " ";
      cout << endl;
    }
  }
  return 0;
}
