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 = 222222;
int a[N], b[N], cnt[N], c[N], d[N], col[N], vis[N];
vector<pair<int, int> > graph[N];
vector<int> comp[N];
int n;
int dfs(int u = 1, int c = 1) {
  if (col[u]) {
    return col[u] == c ? 0 : -1;
  }
  col[u] = c;
  for (auto p : graph[u]) {
    dfs(p.first, p.second ? 3 - c : c);
  }
  return 0;
}
pair<int, int> dfs2(int u, int comp) {
  if (vis[u]) return {0, 0};
  vis[u] = comp;
  int total = 1, sw = col[u] - 1;
  for (auto p : graph[u]) {
    pair<int, int> pp = dfs2(p.first, comp);
    total += pp.first;
    sw += pp.second;
  }
  return {total, sw};
}
int solve() {
  cin >> n;
  fill(c + 1, c + n + 1, 0);
  fill(d + 1, d + n + 1, 0);
  for (int i = (1); i <= (n); i++) {
    cin >> a[i];
    comp[i].clear();
    cnt[i] = 0, col[i] = 0, vis[i] = 0;
    graph[i].clear();
    if (c[a[i]])
      d[a[i]] = i;
    else
      c[a[i]] = i;
  }
  for (int i = (1); i <= (n); i++) {
    cin >> b[i];
    if (c[b[i]])
      d[b[i]] = i;
    else
      c[b[i]] = i;
  }
  for (int i = (1); i <= (n); i++) {
    cnt[a[i]]++;
    cnt[b[i]]++;
  }
  for (int x = (1); x <= (n); x++) {
    if (cnt[x] != 2) {
      cout << -1 << endl;
      return 0;
    }
    if (c[x] != d[x]) {
      bool ch =
          (a[c[x]] == x && a[d[x]] == x) || (b[c[x]] == x && b[d[x]] == x);
      graph[c[x]].push_back({d[x], ch});
      graph[d[x]].push_back({c[x], ch});
    }
  }
  for (int u = (1); u <= (n); u++) {
    if (!col[u]) {
      dfs(u);
    }
  }
  int cn = 0;
  for (int u = (1); u <= (n); u++) {
    if (!vis[u]) {
      dfs2(u, ++cn);
    }
    comp[vis[u]].push_back(u);
  }
  vector<int> ans;
  for (int i = (1); i <= (cn); i++) {
    int change = 0;
    for (auto u : comp[i]) {
      change += col[u] - 1;
    }
    change = change * 2 > (int)(comp[i]).size();
    for (auto u : comp[i]) {
      bool ch = change ^ (col[u] == 2);
      if (ch) {
        ans.push_back(u);
      }
    }
  }
  cout << (int)(ans).size() << endl;
  for (auto u : ans) {
    cout << u << " ";
  }
  cout << endl;
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
