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;
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n), b(n), cnt(n + 1);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      cnt[a[i]]++;
    }
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      cnt[b[i]]++;
    }
    bool flag = true;
    for (int i = 1; i <= n; i++) {
      if (cnt[i] != 2) flag = false;
    }
    if (!flag) {
      cout << -1 << "\n";
      continue;
    }
    vector<vector<pair<int, int>>> graph(n + 1);
    map<pair<int, int>, int> dir;
    for (int i = 0; i < n; i++) {
      graph[a[i]].push_back({b[i], i + 1});
      graph[b[i]].push_back({a[i], i + 1});
      if (dir[{a[i], b[i]}] == 1) {
        dir[{a[i], b[i]}] = 0;
        continue;
      }
      dir[{a[i], b[i]}] = 0;
      dir[{b[i], a[i]}] = 1;
    }
    vector<bool> visit(n + 1, false);
    vector<int> ans;
    for (int i = 1; i <= n; i++) {
      if (!visit[i]) {
        vector<int> ans0, ans1;
        queue<int> q;
        q.push(i);
        int v;
        while (!q.empty()) {
          v = q.front();
          q.pop();
          visit[v] = true;
          for (pair<int, int> to : graph[v]) {
            int nv = to.first;
            int idx = to.second;
            if (!visit[nv]) {
              q.push(nv);
              if (dir[{v, nv}] == 0)
                ans0.push_back(idx);
              else
                ans1.push_back(idx);
              break;
            }
          }
        }
        for (pair<int, int> to : graph[v]) {
          int nv = to.first;
          int idx = to.second;
          if (nv == i) {
            if (dir[{v, i}] == 0)
              ans0.push_back(idx);
            else
              ans1.push_back(idx);
            break;
          }
        }
        if (ans0.size() < ans1.size()) {
          for (int to : ans0) ans.push_back(to);
        } else {
          for (int to : ans1) ans.push_back(to);
        }
      }
    }
    cout << ans.size() << "\n";
    for (int to : ans) cout << to << " ";
    cout << "\n";
  }
}
