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;
using vi = vector<int>;
using pii = pair<int, int>;
struct DSU {
  vector<int> s, flip;
  DSU(int n) : s(n, -1), flip(n) {}
  void init(int i) { flip[i] = 1; }
  int find(int i) { return s[i] < 0 ? i : s[i] = find(s[i]); }
  bool join(int a, int b) {
    a = find(a), b = find(b);
    if (a == b) return false;
    if (s[a] > s[b]) swap(a, b);
    flip[a] += flip[b];
    s[a] += s[b], s[b] = a;
    return true;
  }
  int size(int i) { return -s[find(i)]; }
  int ans(int i) { return flip[find(i)]; }
  bool same(int a, int b) { return find(a) == find(b); }
};
void solve() {
  int n;
  cin >> n;
  vector<vi> a(2, vi(n)), pos(n);
  vi f(n);
  DSU dsu(2 * n);
  for (int j = 0; j < 2; j++)
    for (int i = 0; i < n; i++) {
      cin >> a[j][i];
      a[j][i]--;
      f[a[j][i]]++;
      pos[a[j][i]].push_back(i);
      dsu.init(i + n);
    }
  if (*min_element(f.begin(), f.end()) < 2) {
    cout << -1 << '\n';
    return;
  }
  for (int i = 0; i < n; i++) {
    if (a[0][i] == a[1][i]) continue;
    for (int k : pos[a[0][i]]) {
      if (k == i) continue;
      if (a[0][i] == a[0][k])
        dsu.join(i, k + n), dsu.join(i + n, k);
      else
        dsu.join(i, k), dsu.join(i + n, k + n);
    }
    for (int k : pos[a[1][i]]) {
      if (k == i) continue;
      if (a[1][i] == a[0][k])
        dsu.join(i, k), dsu.join(i + n, k + n);
      else
        dsu.join(i, k + n), dsu.join(i + n, k);
    }
  }
  for (int i = 0; i < n; i++)
    if (dsu.same(i, i + n)) {
      cout << -1 << '\n';
      return;
    }
  int ans = 0;
  set<int> take;
  for (int i = 0; i < n; i++) {
    if (take.count(dsu.find(i)) || take.count(dsu.find(i + n))) continue;
    int here = min(dsu.ans(i), dsu.ans(i + n));
    if (dsu.ans(i) == here)
      take.insert(dsu.find(i));
    else
      take.insert(dsu.find(i + n));
    ans += here;
  }
  cout << ans << '\n';
  for (int i = n; i < 2 * n; i++)
    if (take.count(dsu.find(i))) cout << i + 1 - n << " ";
  cout << '\n';
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
