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;
mt19937 rnd;
const int N = 2e5 + 10;
const vector<int> no = {-1};
int a[3][N], b[N], c[N];
vector<pair<int, bool>> g[N];
bool used[N];
void init(int n) {
  for (int i = 1; i <= n; i++) {
    b[i] = c[i] = 0;
    g[i].clear();
    used[i] = 0;
  }
}
void dfs(int v, vector<bool>& cur, vector<int>& pos) {
  used[v] = 1;
  for (auto e : g[v]) {
    int to = e.first;
    if (!used[to]) {
      cur.push_back(e.second);
      pos.push_back(v);
      dfs(to, cur, pos);
      break;
    }
  }
}
bool f(bool x) { return x ? 0 : 1; }
vector<int> solve(int n) {
  init(n);
  for (int i = 1; i <= 2; i++) {
    for (int j = 1; j <= n; j++) {
      if (!b[a[i][j]]) {
        b[a[i][j]] = j;
      } else if (!c[a[i][j]]) {
        c[a[i][j]] = j;
      } else {
        return no;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (b[i] == c[i]) {
      continue;
    }
    bool good = 0;
    if (a[1][b[i]] == i && a[2][c[i]] == i) {
      good = 1;
    }
    if (a[2][b[i]] == i && a[1][c[i]] == i) {
      good = 1;
    }
    g[b[i]].push_back({c[i], good});
    g[c[i]].push_back({b[i], good});
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    if (!used[i] && g[i].size() == 2) {
      vector<bool> cur;
      vector<int> pos;
      dfs(i, cur, pos);
      cur.push_back(g[i][1].second);
      pos.push_back(g[i][1].first);
      int k = cur.size();
      int cnt = 0;
      for (int i = 0; i < k; i++) {
        cnt += !cur[i];
      }
      if (cnt & 1) {
        return no;
      }
      vector<bool> temp = cur;
      vector<int> x, y;
      for (int i = 1; i < k; i++) {
        if (!temp[i - 1]) {
          x.push_back(pos[i]);
          temp[i - 1] = f(temp[i - 1]);
          temp[i] = f(temp[i]);
        }
      }
      temp = cur;
      y.push_back(pos[0]);
      temp[0] = f(temp[0]);
      temp[k - 1] = f(temp[k - 1]);
      for (int i = 1; i < k; i++) {
        if (!temp[i - 1]) {
          y.push_back(pos[i]);
          temp[i - 1] = f(temp[i - 1]);
          temp[i] = f(temp[i]);
        }
      }
      if (x.size() <= y.size()) {
        for (int i : x) ans.push_back(i);
      } else {
        for (int i : y) ans.push_back(i);
      }
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 1; i <= 2; i++) {
      for (int j = 1; j <= n; j++) {
        cin >> a[i][j];
      }
    }
    vector<int> ans = solve(n);
    if (ans.size() == 1 && ans[0] == -1) {
      cout << "-1\n";
    } else {
      cout << ans.size() << "\n";
      for (int i : ans) {
        cout << i << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}
