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 maxn = 5e5 + 888;
int ori[maxn][2];
vector<int> pos[maxn];
int f[maxn];
vector<int> ans;
void cal(vector<int>& se, int val) {
  int flag = 0;
  if (val == ori[se[0]][0]) {
    val = ori[se[0]][1];
    flag = 1;
  } else
    val = ori[se[0]][0];
  int sta = 0;
  int cnt0 = 1, cnt1 = 0;
  f[0] = 0;
  flag ^= 1;
  for (int i = 1; i < se.size(); i++) {
    int pos = se[i];
    if (ori[pos][flag] == val) {
      f[i] = 0;
      cnt0++;
      val = ori[pos][flag ^ 1];
    } else {
      val = ori[pos][flag];
      f[i] = 1;
      cnt1++;
    }
  }
  int pp;
  if (cnt0 < cnt1)
    pp = 0;
  else
    pp = 1;
  for (int i = 0; i < se.size(); i++)
    if (f[i] == pp) ans.push_back(se[i]);
}
void solve() {
  int n;
  scanf("%d", &n);
  ans.clear();
  for (int i = 0; i <= n + 9; i++) pos[i].clear();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ori[i][0]);
    pos[ori[i][0]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ori[i][1]);
    pos[ori[i][1]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (pos[i].size() != 2) {
      puts("-1");
      return;
    }
  }
  map<int, int> ma;
  vector<int> se;
  for (int val = 1; val <= n; val++) {
    int nowpos = pos[val][0];
    if (pos[val][0] == pos[val][1]) continue;
    if (ma[nowpos]) continue;
    se.push_back(nowpos);
    ma[nowpos] = 1;
    int tempval = val;
    int nexval = ori[nowpos][0];
    if (nexval == tempval) nexval = ori[nowpos][1];
    while (nexval != val) {
      int flag = 0;
      for (int p : pos[nexval]) {
        if (ma.count(p)) continue;
        flag = 1;
        se.push_back(p);
        ma[p] = 1;
        nowpos = p;
      }
      if (!flag) {
        puts("-1");
        return;
      }
      tempval = nexval;
      nexval = ori[nowpos][0];
      if (nexval == tempval) nexval = ori[nowpos][1];
    }
    cal(se, val);
    se.clear();
  }
  cout << ans.size() << endl;
  for (int aa : ans) cout << aa << " ";
  cout << endl;
}
signed main() {
  int T;
  cin >> T;
  while (T--) solve();
}
