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 N = 200005;
int n, m;
int a[2][N];
vector<pair<int, int>> pos[N];
bool vis[N], swp[N];
int solve() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    pos[i].clear();
    vis[i] = false;
    swp[i] = false;
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[0][i]);
    pos[--a[0][i]].emplace_back(0, i);
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[1][i]);
    pos[--a[1][i]].emplace_back(1, i);
  }
  for (int i = 0; i < n; ++i) {
    if (pos[i].size() != 2) return 0 * printf("-1\n");
  }
  vector<int> ans;
  for (int i = 0; i < n; ++i) {
    if (vis[a[0][i]]) continue;
    vector<int> cyc;
    int lst = -1;
    int cur = a[0][i];
    int cnt = 0;
    bool same = false;
    while (!vis[cur]) {
      vis[cur] = true;
      if (pos[cur][0].second == lst) {
        swap(pos[cur][0], pos[cur][1]);
      }
      cyc.push_back(pos[cur][0].second);
      if (pos[cur][0].first == pos[cur][1].first) same = !same;
      if (same) {
        ++cnt;
        swp[pos[cur][0].second] = true;
      }
      lst = pos[cur][0].second;
      cur = a[pos[cur][0].first ^ 1][pos[cur][0].second];
    }
    if (cnt < (int)cyc.size() - cnt) {
      for (int c : cyc) {
        if (swp[c]) ans.push_back(c);
      }
    } else {
      for (int c : cyc) {
        if (!swp[c]) ans.push_back(c);
      }
    }
  }
  printf("%d\n", (int)ans.size());
  for (int u : ans) printf("%d ", u + 1);
  printf("\n");
  return 0;
}
int main() {
  int t = 1;
  scanf("%d", &t);
  for (int tc = 0; tc < t; ++tc) {
    solve();
  }
  return 0;
}
