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 + 10;
int col[maxn];
vector<int> paint[2];
vector<pair<int, int> > graph[maxn];
void dfs(int u) {
  paint[col[u]].push_back(u);
  for (auto e : graph[u]) {
    int v = e.second, w = e.first;
    if (col[v] != -1) continue;
    if (w)
      col[v] = 1 - col[u];
    else
      col[v] = col[u];
    dfs(v);
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    vector<int> a[2], id[n], cnt(n, 0);
    a[0].resize(n), a[1].resize(n);
    for (int i = 0; i < 2; ++i) {
      for (int j = 0; j < n; ++j) {
        cin >> a[i][j];
        a[i][j]--;
        id[a[i][j]].push_back(j);
        cnt[a[i][j]]++;
      }
    }
    bool imp = 1;
    for (int i = 0; i < n; ++i) {
      graph[i].clear();
      col[i] = -1;
      imp &= (cnt[i] == 2);
    }
    if (!imp) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 0; i < n; ++i) {
      int c1 = id[i][0], c2 = id[i][1];
      if (c1 == c2) continue;
      int r1 = a[0][c1] == i ? 0 : 1, r2 = a[0][c2] == i ? 0 : 1;
      int w = (r1 == r2);
      graph[c1].push_back({w, c2});
      graph[c2].push_back({w, c1});
    }
    vector<int> ans;
    for (int i = 0; i < n; ++i) {
      if (graph[i].empty() || col[i] != -1) continue;
      col[i] = 0;
      paint[0].clear(), paint[1].clear();
      dfs(i);
      if (paint[0].size() < paint[1].size()) {
        for (int i : paint[0]) ans.push_back(i + 1);
      } else
        for (int i : paint[1]) ans.push_back(i + 1);
    }
    cout << ans.size() << endl;
    for (int i : ans) cout << i << ' ';
    cout << endl;
  }
  return 0;
}
