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 = 2e5 + 100;
int n;
int t[2][N];
vector<int> vals[N];
vector<pair<int, int>> G[N];
bool vis[N];
int color[N];
vector<int> C[2];
bool cycle;
void dfs(int v) {
  vis[v] = true;
  C[color[v]].push_back(v);
  for (auto it : G[v]) {
    if (vis[it.first] && (color[v] ^ it.second) != color[it.first])
      cycle = true;
    if (!vis[it.first]) {
      color[it.first] = color[v] ^ it.second;
      dfs(it.first);
    }
  }
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) G[i].clear(), vals[i].clear();
  for (int j = 0; j <= 1; ++j)
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &t[j][i]);
      vals[t[j][i]].push_back(i);
    }
  for (int i = 1; i <= n; ++i) {
    if ((int)vals[i].size() != 2) {
      printf("-1\n");
      return;
    }
    int u = vals[i][0], v = vals[i][1], w;
    if (u == v) continue;
    w = ((t[0][u] == t[0][v] && t[0][u] == i) ||
         (t[1][u] == t[1][v] && t[1][u] == i));
    G[u].push_back(make_pair(v, w));
    G[v].push_back(make_pair(u, w));
  }
  cycle = false;
  for (int i = 1; i <= n; ++i) vis[i] = false;
  vector<int> ans;
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) {
      for (int j = 0; j <= 1; ++j) C[j].clear();
      dfs(i);
      if ((int)C[0].size() > (int)C[1].size()) swap(C[0], C[1]);
      for (auto it : C[0]) ans.push_back(it);
    }
  if (cycle) printf("-1\n");
  printf("%d\n", (int)ans.size());
  for (auto it : ans) printf("%d ", it);
  printf("\n");
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    solve();
  }
  return 0;
}
