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>
struct UF {
  std::vector<int> e;
  UF(int n) : e(n, -1) {}
  int find(int x) { return e[x] < 0 ? x : e[x] = find(e[x]); }
  bool join(int a, int b) {
    a = find(a), b = find(b);
    if (a == b) return false;
    if (e[a] > e[b]) std::swap(a, b);
    e[a] += e[b];
    e[b] = a;
    return true;
  }
};
const int N = 400040;
std::pair<int, int> a[N];
int _, n, flag, d[N], s[2][N];
int main() {
  for (scanf("%d", &_); _--; flag = 0) {
    scanf("%d", &n);
    UF uf(2 * n);
    for (int i = 0; i < 2 * n; i++) scanf("%d", &a[i].first), a[i].second = i;
    sort(a, a + 2 * n);
    for (int i = 0; i < 2 * n; i += 2) {
      int x = a[i].second;
      int y = a[i + 1].second;
      if (a[i].first != i / 2 + 1) flag = 1;
      if (a[i + 1].first != i / 2 + 1) flag = 1;
      if (x / n == y / n)
        uf.join(x % n, y % n + n), uf.join(x % n + n, y % n);
      else
        uf.join(x % n, y % n), uf.join(x % n + n, y % n + n);
    }
    for (int i = 0; i < 2 * n; i++) s[i < n][uf.find(i)]++;
    for (int i = 0; i < n; i++) {
      int fi = uf.find(i);
      int se = uf.find(i + n);
      if (s[0][fi] < s[0][se])
        d[fi] = 0, d[se] = 1;
      else
        d[fi] = 1, d[se] = 0;
    }
    std::vector<int> ans;
    for (int i = 0; i < n; i++)
      if (d[uf.find(i)]) ans.push_back(i + 1);
    if (!flag) {
      printf("%d\n", ans.size());
      for (int j : ans) printf("%d ", j);
      puts("");
    } else
      puts("-1");
    for (int i = 0; i < 2 * n; i++) s[1][i] = s[0][i] = d[i] = 0;
  }
}
