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 = 1e6;
int a[N + 2], b[N + 2], fr[N + 2], vis[N + 2], cl[N + 2];
vector<int> pa[N + 2], pb[N + 2], bf;
vector<pair<int, int> > adj[N + 2];
int f;
void dfs(int node, int c) {
  if (vis[node]) {
    if (cl[node] != c) f = 1;
    return;
  }
  bf.push_back(node);
  cl[node] = c;
  vis[node] = 1;
  for (auto x : adj[node]) dfs(x.first, c ^ x.second);
}
vector<int> ans;
bool solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++)
    pa[i].clear(), pb[i].clear(), adj[i].clear(), fr[i] = 0, vis[i] = 0;
  f = 0;
  ans.clear();
  for (int i = 1; i <= n; i++) cin >> a[i], fr[a[i]]++, pa[a[i]].push_back(i);
  for (int i = 1; i <= n; i++) cin >> b[i], fr[b[i]]++, pb[b[i]].push_back(i);
  for (int i = 1; i <= n; i++)
    if (fr[i] != 2) return false;
  for (int i = 1; i <= n; i++) {
    if (pa[i].size() == 2) {
      adj[pa[i][0]].push_back({pa[i][1], 1});
      adj[pa[i][1]].push_back({pa[i][0], 1});
    } else if (pb[i].size() == 2) {
      adj[pb[i][0]].push_back({pb[i][1], 1});
      adj[pb[i][1]].push_back({pb[i][0], 1});
    } else {
      adj[pa[i][0]].push_back({pb[i][0], 0});
      adj[pb[i][0]].push_back({pa[i][0], 0});
    }
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    bf.clear();
    dfs(i, 0);
    if (f) return false;
    int cnt = 0;
    for (auto x : bf) cnt += cl[x];
    int c = 0;
    if (cnt < bf.size() - cnt) c = 1;
    for (auto x : bf)
      if (cl[x] == c) ans.push_back(x);
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    if (solve()) {
      cout << ans.size() << '\n';
      for (auto x : ans) cout << x << " ";
      cout << '\n';
    } else
      cout << -1 << '\n';
  }
  return 0;
}
