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;
int t, n, ans;
vector<int> num1, num2, used, res;
vector<vector<int>> pos;
int other(int p, int x) {
  if (num1[p] == x) return num2[p];
  return num1[p];
}
int otherpos(int x, int p) {
  for (int i = 0; i < pos[x].size(); i++)
    if (pos[x][i] != p) return pos[x][i];
}
void rfs(int x) {
  int st, now, opp;
  vector<int> seq, compare, where;
  st = now = num1[x];
  used[now] = 1;
  compare.push_back(st);
  seq.push_back(st);
  where.push_back(x);
  opp = other(x, now);
  while (opp != st) {
    x = otherpos(opp, x);
    now = opp;
    opp = other(x, now);
    used[now] = 1;
    compare.push_back(now);
    seq.push_back(num1[x]);
    where.push_back(x);
  }
  int m1, m2, s;
  m1 = m2 = 0;
  s = seq.size();
  for (int i = 0; i < s; i++)
    if (seq[i] != compare[i]) ++m1;
  for (int i = 0; i < s; i++)
    if (seq[i] != compare[(i + 1) % s]) ++m2;
  if (m1 < m2) {
    ans += m1;
    for (int i = 0; i < s; i++)
      if (seq[i] != compare[i]) res.push_back(where[i] + 1);
  } else {
    ans += m2;
    for (int i = 0; i < s; i++)
      if (seq[i] != compare[(i + 1) % s]) res.push_back(where[i] + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    num1 = num2 = used = vector<int>(n);
    pos = vector<vector<int>>(n, vector<int>());
    ans = 0;
    res.clear();
    int x;
    for (int i = 0; i < n; i++) {
      cin >> x;
      --x;
      num1[i] = x;
      pos[x].push_back(i);
    }
    for (int i = 0; i < n; i++) {
      cin >> x;
      --x;
      num2[i] = x;
      pos[x].push_back(i);
    }
    bool bad = false;
    for (int i = 0; i < n; i++) {
      if (pos[i].size() != 2) {
        bad = true;
        break;
      }
    }
    if (bad) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 0; i < n; i++) {
      if (used[num1[i]] == 0) rfs(i);
    }
    cout << ans << endl;
    for (int i = 0, j = res.size(); i < j; i++) cout << res[i] << " ";
    cout << endl;
  }
  return 0;
}
