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;
template <class t>
inline t read(t &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const int N = 2e5 + 5;
int n, v[N], col[N], fl;
vector<int> ans, p[2];
vector<pair<int, int>> a[N], g[N];
void dfs(int x, int c) {
  v[x] = 1;
  col[x] = c;
  p[c].push_back(x);
  for (auto o : g[x]) {
    int y = o.first, w = o.second;
    if (!v[y])
      dfs(y, c ^ w);
    else if (col[y] ^ c ^ w)
      fl = 0;
  }
}
void doit() {
  read(n);
  ans.clear();
  for (int i = 1; i <= n; i++) a[i].clear(), g[i].clear(), v[i] = 0;
  for (int o : {0, 1})
    for (int i = 1, x; i <= n; i++) a[read(x)].emplace_back(o, i);
  for (int i = 1; i <= n; i++) {
    if (a[i].size() < 2) return puts("-1"), void();
    int x = a[i][0].second, y = a[i][1].second,
        w = a[i][0].first == a[i][1].first;
    g[x].emplace_back(y, w), g[y].emplace_back(x, w);
  }
  fl = 1;
  for (int i = 1; i <= n; i++)
    if (!v[i]) {
      p[0].clear();
      p[1].clear();
      dfs(i, 0);
      if (p[0].size() > p[1].size()) swap(p[0], p[1]);
      for (int x : p[0]) ans.push_back(x);
    }
  if (!fl) return puts("-1"), void();
  write(ans.size());
  puts("");
  for (int x : ans) write(x), putchar(' ');
  puts("");
}
signed main() {
  int t;
  read(t);
  while (t--) doit();
}
