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 const mod = 1e9 + 7;
long long const inf = 2e18;
inline long long add(long long a, long long b) {
  return (a % mod + b % mod + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return (a % mod * b % mod + mod) % mod;
}
inline long long sub(long long a, long long b) {
  return (a % mod - b % mod + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long modexp(long long x, long long n) {
  if (n == 0) {
    return 1;
  } else if (n & 1) {
    return (modexp((x * x) % mod, n / 2) % mod * x % mod) % mod;
  } else {
    return (modexp((x * x) % mod, n / 2) % mod);
  }
}
long long modinv(long long n) { return modexp(n, mod - 2); }
long long fn(long long v, long long idx, set<long long> second[]) {
  for (auto it : second[v]) {
    if (it != idx) return it;
  }
}
void solve() {
  long long n;
  cin >> n;
  long long a[2 * n];
  set<long long> second[n + 1];
  for (long long i = 0; i < (2 * n); i++) {
    cin >> a[i];
    second[a[i]].insert(i);
  }
  for (int i = 1; i <= n; i++) {
    if (second[i].size() != 2) {
      cout << "-1\n";
      return;
    }
  }
  bool vis[n];
  memset(vis, false, sizeof(vis));
  vector<long long> ans;
  for (int i = 0; i < n; i++) {
    if (vis[i]) continue;
    if (a[i] == a[i + n]) {
      vis[i] = true;
      continue;
    }
    long long k = 0;
    vis[i] = true;
    long long v = a[i];
    long long idx = i;
    long long pos = 0;
    vector<pair<long long, long long> > temp;
    temp.push_back({i, 0});
    while (1) {
      idx = fn(v, idx, second);
      if (vis[idx % n]) break;
      vis[idx % n] = true;
      if (idx / n == pos) {
        temp.push_back({idx % n, 1});
        k++;
        pos = idx / n;
        idx = (idx + n) % (2 * n);
        v = a[idx];
      } else {
        temp.push_back({idx % n, 0});
        idx = (idx + n) % (2 * n);
        pos = idx / n;
        v = a[idx];
      }
    }
    if (k < temp.size() - k) {
      for (auto it : temp) {
        if (it.second == 1) ans.push_back(it.first + 1);
      }
    } else {
      for (auto it : temp) {
        if (it.second == 0) ans.push_back(it.first + 1);
      }
    }
  }
  cout << ans.size() << "\n";
  for (auto it : ans) {
    cout << it << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T = 1;
  cin >> T;
  for (int t = 1; t <= T; t++) {
    solve();
  }
  return 0;
}
