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 long long N = 2e5 + 10;
long long vis[N];
vector<long long> adj[N][2];
long long cnt, ones;
long long mp[N];
pair<long long, long long> lst[N];
long long a[N][2];
vector<long long> stk;
long long finalAns[N];
void dfs(long long i, long long mode) {
  stk.push_back(i);
  vis[i] = 1;
  finalAns[i] = mode;
  cnt++;
  if (mode) ones++;
  for (long long c = 0; c < (long long)2; c++) {
    for (long long v : adj[i][c]) {
      if (!vis[v]) dfs(v, mode ^ c);
    }
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tt;
  cin >> tt;
  while (tt--) {
    long long n;
    cin >> n;
    for (long long i = 0; i < (long long)n + 2; i++) {
      for (long long first = 0; first < (long long)2; first++)
        adj[i][first].clear();
      vis[i] = 0;
      mp[i] = 0;
      lst[i].first = -1;
    }
    for (long long i = 0; i < (long long)n; i++) cin >> a[i][0], mp[a[i][0]]++;
    for (long long i = 0; i < (long long)n; i++) cin >> a[i][1], mp[a[i][1]]++;
    bool fail = 0;
    for (long long i = 1; i <= n; i++) {
      if (mp[i] != 2) fail = 1;
    }
    if (fail) {
      cout << -1 << "\n";
      continue;
    }
    for (long long e = 0; e < (long long)n; e++) {
      for (long long r = 0; r < (long long)2; r++) {
        long long i = a[e][r];
        if (lst[i].first == -1) {
          lst[i] = {r, e};
        } else {
          if (lst[i].first == r) {
            adj[e][1].push_back(lst[i].second);
            adj[lst[i].second][1].push_back(e);
          } else {
            adj[e][0].push_back(lst[i].second);
            adj[lst[i].second][0].push_back(e);
          }
        }
      }
    }
    long long ans = 0;
    for (long long i = 0; i < n; i++) {
      if (!vis[i]) {
        cnt = ones = 0;
        stk.clear();
        dfs(i, 0);
        if (ones > cnt - ones) {
          for (long long i : stk) finalAns[i] ^= 1;
        }
      }
    }
    long long ccc = 0;
    for (long long i = 0; i < (long long)n; i++) ccc += finalAns[i];
    cout << ccc << "\n";
    for (long long i = 0; i < (long long)n; i++) {
      if (finalAns[i]) cout << i + 1 << " ";
    }
    cout << "\n";
  }
}
