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 max_n = 4e5 + 5;
int t, n, a, ciag[max_n];
vector<int> wystapienia[max_n], graf[max_n], ile, ile_2, wynik;
bool odwiedzony[max_n];
void wczytaj() {
  cin >> n;
  for (int i = 1; i <= 2 * n; i++) {
    cin >> ciag[i];
    wystapienia[ciag[i]].push_back(i);
  }
}
bool sprawdz() {
  for (int i = 1; i <= n; i++)
    if (wystapienia[i].size() != 2) return false;
  return true;
}
void stworz_graf() {
  for (int i = 1; i <= n; i++) {
    int pom;
    if (i + n == wystapienia[ciag[i + n]][0])
      pom = wystapienia[ciag[i + n]][1];
    else
      pom = wystapienia[ciag[i + n]][0];
    graf[i].push_back(pom);
    graf[pom].push_back(i);
  }
  for (int i = 1 + n; i <= 2 * n; i++) {
    int pom;
    if (i - n == wystapienia[ciag[i - n]][0])
      pom = wystapienia[ciag[i - n]][1];
    else
      pom = wystapienia[ciag[i - n]][0];
    graf[i].push_back(pom);
    graf[pom].push_back(i);
  }
}
void dfs(int x) {
  if (x > n)
    ile.push_back(x - n);
  else
    ile_2.push_back(x);
  for (int i : graf[x]) {
    if (i > n && !odwiedzony[i - n]) {
      odwiedzony[i - n] = 1;
      dfs(i);
    }
    if (i <= n && !odwiedzony[i]) {
      odwiedzony[i] = 1;
      dfs(i);
    }
  }
}
void solve() {
  for (int i = 1; i <= n; i++)
    if (!odwiedzony[i]) {
      ile.clear();
      ile_2.clear();
      odwiedzony[i] = 1;
      dfs(i);
      if (ile.size() > ile_2.size()) ile = ile_2;
      for (int j : ile) wynik.push_back(j);
    }
  cout << wynik.size() << "\n";
  for (int i : wynik) cout << i << " ";
  cout << "\n";
}
void reset() {
  wynik.clear();
  for (int i = 1; i <= 2 * n; i++) {
    wystapienia[i].clear();
    graf[i].clear();
    odwiedzony[i] = 0;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    wczytaj();
    if (!sprawdz()) {
      cout << "-1\n";
      reset();
      continue;
    }
    stworz_graf();
    solve();
    reset();
  }
  return 0;
}
