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;
vector<int> r1, r2;
vector<bool> pd;
vector<int> out;
vector<pair<int, int>> mp;
int le(int i, int cur) {
  if (cur == mp[i].first) return mp[i].second;
  return mp[i].first;
}
bool mak(int i, int p) {
  if (mp[i].first == -1)
    mp[i].first = p;
  else if (mp[i].second == -1)
    mp[i].second = p;
  else
    return true;
  return false;
}
class proc {
  struct Dt {
    int pos;
    bool fl;
  };
  int p;
  int c;
  vector<Dt> out_c;

 public:
  proc(int i) {
    p = i;
    c = 0;
    re();
    if (c > out_c.size() / 2) {
      for (auto& i : out_c)
        if (!i.fl) out.push_back(i.pos);
    } else
      for (auto& i : out_c)
        if (i.fl) out.push_back(i.pos);
  }

 private:
  void re() {
    if (pd[p]) return;
    pd[p] = 1;
    if (r1[p] == r2[p]) return;
    int x = le(r1[p], p);
    if (r1[p] == r1[x]) {
      ++c;
      swap(r1[x], r2[x]);
      out_c.push_back({x, 1});
    } else
      out_c.push_back({x, 0});
    p = x;
    re();
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    r1.resize(n);
    r2.resize(n);
    mp = vector<pair<int, int>>(n, {-1, -1});
    pd = vector<bool>(n, 0);
    out = vector<int>(0);
    for (int i = 0; i < n; ++i) {
      cin >> r1[i];
      --r1[i];
    }
    for (int i = 0; i < n; ++i) {
      cin >> r2[i];
      --r2[i];
    }
    for (int i = 0; i < n; ++i)
      if (mak(r1[i], i) || mak(r2[i], i)) goto badcase;
    for (int i = 0; i < n; ++i) proc lp(i);
    cout << out.size() << endl;
    for (auto i : out) cout << i + 1 << ' ';
    cout << endl;
    continue;
  badcase:
    cout << -1 << endl;
  }
}
