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 t, n, v;
long long a[200005], b[200005], vis[200005], cnt[200005];
vector<long long> g[200005];
vector<long long> ans, top, bot;
void dfs(long long u, long long pos) {
  if (vis[u]) return;
  vis[u] = 1;
  for (long long p : g[u]) {
    if (p != pos) {
      if (a[p] == u)
        v = b[p], top.push_back(p);
      else
        v = a[p], bot.push_back(p);
      dfs(v, p);
      return;
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> t;
  while (t--) {
    cin >> n;
    for (long long i = 1; i <= n; i++) {
      g[i].clear();
      vis[i] = 0;
      cnt[i] = 0;
    }
    ans.clear();
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
      cnt[a[i]]++;
      g[a[i]].push_back(i);
    }
    for (long long i = 1; i <= n; i++) {
      cin >> b[i];
      cnt[b[i]]++;
      g[b[i]].push_back(i);
    }
    long long flag = 0;
    for (long long i = 1; i <= n; i++) {
      if (cnt[i] != 2) {
        flag = 1;
        break;
      }
    }
    if (flag) {
      cout << -1 << endl;
      continue;
    }
    for (long long i = 1; i <= n; i++) {
      if (vis[i]) continue;
      top.clear();
      bot.clear();
      dfs(i, 0);
      if (top.size() < bot.size())
        for (auto x : top) ans.push_back(x);
      else
        for (auto x : bot) ans.push_back(x);
    }
    cout << ans.size() << endl;
    for (auto x : ans) cout << x << ' ';
    cout << endl;
  }
}
