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 nax = 2e5 + 5;
int n, m;
vector<int> adj[nax * 2];
int a[3][nax];
int cnt[nax];
vector<pair<int, int> > pos[nax];
bool vis[nax * 2];
vector<int> ans;
bool ok[nax];
int all;
int taken;
vector<int> cur;
void dfs(int v, int czy) {
  cur.push_back(v);
  if (v <= n) all++;
  vis[v] = true;
  if (czy == 1) {
    if (v <= n) {
      taken++;
      ok[v] = true;
    }
  }
  for (int ziom : adj[v]) {
    if (vis[ziom] == false) dfs(ziom, czy ^ 1);
  }
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    adj[i].clear();
    pos[i].clear();
    vis[i] = false;
    ok[i] = false;
  }
  for (int i = n + 1; i <= 2 * n; i++) adj[i].clear();
  for (int i = n + 1; i <= 2 * n; i++) vis[i] = false;
  ans.clear();
  for (int i = 1; i <= n; i++) cnt[i] = 0;
  for (int i = 1; i <= 2; i++) {
    for (int j = 1; j <= n; j++) cin >> a[i][j];
  }
  for (int i = 1; i <= 2; i++) {
    for (int j = 1; j <= n; j++) cnt[a[i][j]]++;
  }
  for (int i = 1; i <= n; i++) {
    if (cnt[i] != 2) {
      cout << "-1" << endl;
      return;
    }
  }
  for (int i = 1; i <= 2; i++) {
    for (int j = 1; j <= n; j++) {
      int akt = a[i][j];
      pos[akt].push_back({i, j});
    }
  }
  int cnt = n + 1;
  for (int i = 1; i <= n; i++) {
    if (pos[i].size() == 2) {
      if (pos[i][0].second == pos[i][1].second) continue;
      if (pos[i][0].first == pos[i][1].first) {
        int x = pos[i][0].second;
        int y = pos[i][1].second;
        adj[x].push_back(y);
        adj[y].push_back(x);
      } else {
        int x = pos[i][0].second;
        int y = pos[i][1].second;
        adj[cnt].push_back(x);
        adj[cnt].push_back(y);
        adj[x].push_back(cnt);
        adj[y].push_back(cnt);
        cnt++;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i] == false && adj[i].size()) {
      cur.clear();
      all = 0;
      taken = 0;
      dfs(i, 1);
      bool flag = 1;
      if (taken * 2 > all) flag = 0;
      for (int s : cur) {
        if (s <= n && ok[s] == flag) ans.push_back(s);
      }
    }
  }
  cout << ans.size() << "\n";
  for (int i : ans) cout << i << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
