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 = 4e5 + 10, mod = 1e9 + 7;
int cmp, n, comp[N];
vector<int> g[N];
bool mark[N];
vector<int> choose, nchoose, topol;
void addedge(int u, int v) {
  g[u].push_back(v);
  g[v].push_back(u);
}
void dfs(int v, int tp) {
  mark[v] = true;
  comp[v] = cmp;
  if (v < n && !mark[v + n]) choose.push_back(v);
  if (v >= n && !mark[v - n]) nchoose.push_back(v - n);
  for (int u : g[v])
    if (!mark[u]) dfs(u, tp);
  if (!tp) topol.push_back(v);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<pair<int, int> > pos[n + 1];
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      pos[x].push_back({0, i});
    }
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      pos[x].push_back({1, i});
    }
    bool bad = false;
    for (int i = 1; i <= n; i++) {
      if (pos[i].size() != 2) {
        bad = true;
      }
    }
    if (bad) {
      cout << "-1\n";
      continue;
    }
    for (int i = 1; i <= n; i++) {
      int c1 = pos[i][0].second, c2 = pos[i][1].second;
      if (pos[i][0].first == pos[i][1].first) {
        addedge(c1, c2 + n);
        addedge(c2, c1 + n);
      } else {
        addedge(c1, c2);
        addedge(c1 + n, c2 + n);
      }
    }
    for (int i = 0; i < 2 * n; i++)
      if (!mark[i]) {
        dfs(i, 0);
        cmp++;
      }
    reverse(topol.begin(), topol.end());
    cmp = 0;
    for (int i = 0; i < 2 * n; i++) mark[i] = false;
    vector<int> ans;
    for (int v : topol)
      if (!mark[v]) {
        choose.clear();
        nchoose.clear();
        dfs(v, 1);
        cmp++;
        if (choose.size() > nchoose.size()) choose.swap(nchoose);
        for (int v : choose) ans.push_back(v);
      }
    for (int i = 0; i < n; i++) {
      if (comp[i] == comp[i + n]) bad = true;
    }
    if (bad)
      cout << "-1\n";
    else {
      cout << ans.size() << "\n";
      for (int v : ans) cout << v + 1 << " ";
      cout << "\n";
    }
    cmp = 0;
    topol.clear();
    for (int i = 0; i < 2 * n; i++) {
      g[i].clear();
      mark[i] = false;
    }
    choose.clear();
    nchoose.clear();
  }
  return 0;
}
