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 + 7;
vector<int> lst[N], pos[N], adj[N];
int col[N];
struct DSU {
  vector<int> p, sz;
  vector<vector<int> > lst;
  DSU(int n) {
    p.resize(n + 1);
    sz.resize(n + 1, 1);
    lst.resize(n + 1);
    for (int i = 1; i <= n; i++) {
      p[i] = i;
      lst[i].push_back(i);
    }
  }
  int find(int x) { return p[x] = (p[x] == x) ? p[x] : find(p[x]); }
  int merge(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) {
      return 0;
    }
    if (sz[x] > sz[y]) {
      swap(x, y);
    }
    for (auto z : lst[x]) {
      lst[y].push_back(z);
    }
    p[x] = y;
    sz[y] += sz[x];
    return 1;
  }
};
vector<int> vec[2];
int dfs(int v, int c, DSU &d) {
  if (~col[v]) {
    if (col[v] ^ c) {
      return 0;
    }
    return 1;
  }
  for (auto k : d.lst[v]) {
    vec[c].push_back(k);
  }
  col[v] = c;
  for (int u : adj[v]) {
    if (!dfs(u, c ^ 1, d)) {
      return 0;
    }
  }
  return 1;
}
void solve() {
  int n;
  cin >> n;
  DSU d(n);
  for (int i = 1; i <= n; i++) {
    lst[i].clear();
    pos[i].clear();
    adj[i].clear();
    col[i] = -1;
  }
  for (int i = 1; i <= 2; i++) {
    for (int j = 1; j <= n; j++) {
      int x;
      cin >> x;
      pos[x].push_back(j);
      lst[x].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++)
    if (lst[i].size() != 2) {
      cout << -1 << endl;
      return;
    }
  for (int i = 1; i <= n; i++) {
    int x = lst[i][0], y = lst[i][1];
    if (x ^ y) {
      d.merge(pos[i][0], pos[i][1]);
    }
  }
  for (int i = 1; i <= n; i++) {
    int x = lst[i][0], y = lst[i][1];
    if (x == y) {
      int p = d.find(pos[i][0]), q = d.find(pos[i][1]);
      if (p == q) {
        cout << -1 << endl;
        return;
      } else {
        adj[p].push_back(q);
        adj[q].push_back(p);
      }
    }
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    int j = d.find(i);
    if (col[j] == -1) {
      vec[0].clear();
      vec[1].clear();
      if (!dfs(j, 0, d)) {
        cout << -1 << endl;
        return;
      }
      if (vec[0].size() > vec[1].size()) {
        swap(vec[0], vec[1]);
      }
      for (int k : vec[0]) {
        ans.push_back(k);
      }
    }
  }
  cout << ans.size() << endl;
  for (int k : ans) {
    cout << k << " ";
  }
  cout << endl;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int test, tc = 0;
  cin >> test;
  while (test--) {
    solve();
  }
}
