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 = 2e5 + 100;
int n;
int a[maxn];
int b[maxn];
int visit[maxn];
vector<int> g[maxn];
vector<int> wjm;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) g[i].clear();
    wjm.clear();
    memset(visit, 0, sizeof(visit));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      g[a[i]].push_back(i);
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d", &b[i]);
      g[b[i]].push_back(i);
    }
    bool f = true;
    for (int i = 1; i <= n; i++) {
      if (g[i].size() != 2) f = false;
    }
    if (!f) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      if (visit[i]) continue;
      vector<int> x, y;
      for (int j = i, k = g[i][0]; !visit[j];) {
        visit[j] = 1;
        k = g[j][0] ^ g[j][1] ^ k;
        if (a[k] == j) {
          x.push_back(k);
          j = b[k];
        } else {
          y.push_back(k);
          j = a[k];
        }
      }
      if (x.size() < y.size())
        wjm.insert(wjm.end(), x.begin(), x.end());
      else
        wjm.insert(wjm.end(), y.begin(), y.end());
    }
    printf("%d\n", wjm.size());
    for (int i = 0; i < wjm.size(); i++) printf("%d ", wjm[i]);
    printf("\n");
  }
  return 0;
}
