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;
using ll = long long;
vector<int> v[400010];
bool vis[400010];
void add(int v1, int v2) {
  v[v1].push_back(v2);
  v[v2].push_back(v1);
}
void dfs(int i, vector<int> &temp) {
  if (vis[i] || v[i].size() == 0) return;
  vis[i] = 1;
  if (i % 2 == 0) temp.push_back(i / 2);
  for (int x : v[i]) dfs(x, temp);
  return;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 0; i < 2 * n + 5; i++) {
      ::v[i].clear();
      vis[i] = 0;
    }
    vector<vector<int>> v(2, vector<int>(n));
    vector<vector<int>> row(n, vector<int>(2, -1)), col = row;
    vector<int> occ(n, 0);
    bool ok = 1;
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < n; j++) {
        scanf("%d", &v[i][j]);
        v[i][j]--;
        occ[v[i][j]]++;
        if (occ[v[i][j]] > 2) ok = 0;
        if (row[v[i][j]][0] == -1)
          row[v[i][j]][0] = i;
        else
          row[v[i][j]][1] = i;
        if (col[v[i][j]][0] == -1)
          col[v[i][j]][0] = j;
        else
          col[v[i][j]][1] = j;
      }
    }
    if (!ok) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 0; i < n; i++) {
      if (col[i][0] == col[i][1]) continue;
      if (row[i][0] == row[i][1]) {
        add(2 * col[i][0], 2 * col[i][1] + 1);
        add(2 * col[i][0] + 1, 2 * col[i][1]);
      } else {
        add(2 * col[i][0], 2 * col[i][1]);
        add(2 * col[i][0] + 1, 2 * col[i][1] + 1);
      }
    }
    vector<int> ans;
    for (int i = 0; i < 2 * n; i += 2) {
      vector<int> temp, temp1;
      dfs(i, temp);
      dfs(i + 1, temp1);
      if (temp1.size() < temp.size()) swap(temp, temp1);
      for (int x : temp) ans.push_back(x);
    }
    cout << ans.size() << endl;
    for (int i : ans) cout << i + 1 << " ";
    cout << endl;
  }
  return 0;
}
