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 = 2e5 + 10;
vector<pair<int, int> > g[N];
int a[N][2];
pair<int, int> pos[N][2];
int vis[N];
vector<int> go[2];
void dfs(int node, int state = 0) {
  vis[node] = 1;
  go[state].push_back(node);
  for (auto x : g[node])
    if (vis[x.second] == 0) dfs(x.second, state ^ x.first);
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) g[i].clear(), vis[i] = 0;
  for (int i = 1; i <= n; i++) cin >> a[i][0];
  for (int i = 1; i <= n; i++) cin >> a[i][1];
  for (int i = 1; i <= n; i++) pos[i][0] = pos[i][1] = {-1, -1};
  int flag = 1;
  for (int i = 1; i <= n; i++) {
    if (pos[a[i][0]][0].first == -1)
      pos[a[i][0]][0] = {i, 0};
    else if (pos[a[i][0]][1].first == -1)
      pos[a[i][0]][1] = {i, 0};
    else
      flag = 0;
    if (pos[a[i][1]][0].first == -1)
      pos[a[i][1]][0] = {i, 1};
    else if (pos[a[i][1]][1].first == -1)
      pos[a[i][1]][1] = {i, 1};
    else
      flag = 0;
  }
  if (flag == 0) {
    cout << -1 << '\n';
    return;
  }
  for (int i = 1; i <= n; i++)
    if (pos[i][0].first != pos[i][1].first) {
      if (pos[i][0].second == pos[i][1].second) {
        g[pos[i][0].first].push_back({1, pos[i][1].first});
        g[pos[i][1].first].push_back({1, pos[i][0].first});
      } else {
        g[pos[i][0].first].push_back({0, pos[i][1].first});
        g[pos[i][1].first].push_back({0, pos[i][0].first});
      }
    }
  vector<int> ans;
  for (int i = 1; i <= n; i++)
    if (vis[i] == 0) {
      dfs(i);
      if (go[0].size() < go[1].size())
        for (auto x : go[0]) ans.push_back(x);
      else
        for (auto x : go[1]) ans.push_back(x);
      go[0].clear();
      go[1].clear();
    }
  cout << ans.size() << '\n';
  for (auto x : ans) cout << x << ' ';
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
}
