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_MAX = 200002;
int t;
int n;
int a[N_MAX], b[N_MAX];
int cnt[N_MAX];
pair<int, bool> x[N_MAX], y[N_MAX];
vector<int> edges[N_MAX][2];
bool color[N_MAX];
bool visited[N_MAX];
vector<int> aux;
void dfs(int u) {
  aux.push_back(u);
  visited[u] = true;
  for (int v : edges[u][0])
    if (visited[v] == false) {
      color[v] = color[u];
      dfs(v);
    }
  for (int v : edges[u][1])
    if (visited[v] == false) {
      color[v] = !color[u];
      dfs(v);
    }
}
void freset() {
  for (int i = 1; i <= n; i++) {
    edges[i][0].clear();
    edges[i][1].clear();
  }
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i <= n; i++) x[i] = y[i] = make_pair(0, 0);
  memset(visited, false, sizeof(visited));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      cnt[a[i]]++;
    }
    for (int i = 1; i <= n; i++) {
      cin >> b[i];
      cnt[b[i]]++;
    }
    bool ok = true;
    for (int i = 1; i <= n; i++)
      if (cnt[i] != 2) ok = false;
    if (ok == false) {
      cout << "-1\n";
      freset();
      continue;
    }
    for (int i = 1; i <= n; i++) {
      if (x[a[i]] == make_pair(0, false))
        x[a[i]] = make_pair(i, false);
      else
        y[a[i]] = make_pair(i, false);
      if (x[b[i]] == make_pair(0, false))
        x[b[i]] = make_pair(i, true);
      else
        y[b[i]] = make_pair(i, true);
    }
    for (int i = 1; i <= n; i++)
      if (x[i].first != y[i].first) {
        bool b = !(x[i].second ^ y[i].second);
        edges[x[i].first][b].push_back(y[i].first);
        edges[y[i].first][b].push_back(x[i].first);
      }
    int ans = 0;
    for (int i = 1; i <= n; i++)
      if (visited[i] == false) {
        dfs(i);
        int cnt[2];
        cnt[0] = cnt[1] = 0;
        for (int u : aux) cnt[color[u]]++;
        if (cnt[0] < cnt[1]) {
          for (int u : aux) color[u] ^= 1;
        }
        ans += min(cnt[0], cnt[1]);
        aux.clear();
      }
    for (int i = 1; i <= n; i++) {
      for (int j : edges[i][0])
        if ((color[i] ^ color[j]) != 0) ok = false;
      for (int j : edges[i][1])
        if ((color[i] ^ color[j]) != 1) ok = false;
    }
    if (ok == false) {
      cout << "-1\n";
      freset();
      continue;
    }
    cout << ans << "\n";
    for (int i = 1; i <= n; i++)
      if (color[i] == true) cout << i << " ";
    cout << "\n";
    freset();
  }
  return 0;
}
