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 n, q;
vector<pair<int, int> > v[maxn];
vector<int> ans, tmp;
bool mark[maxn];
int cnt[maxn], a[maxn], b[maxn];
void dfs(int i) {
  tmp.push_back(i);
  mark[i] = 1;
  for (auto A : v[i])
    if (!mark[A.first]) dfs(A.first);
}
void solve2() {
  if (tmp.size() == 1) return;
  if (tmp.size() == 2) {
    int x = tmp[0];
    if (!((v[x][0].second > n) ^ (v[x][1].second > n)))
      ans.push_back((v[x][0].second - 1) % n + 1);
    return;
  }
  int hel = 0;
  for (int y = 0; y < tmp.size(); y++) {
    if (v[tmp[y]][1].first != tmp[(y + 1) % tmp.size()])
      swap(v[tmp[y]][0], v[tmp[y]][1]);
    hel += (v[tmp[y]][1].second > n);
  }
  if (!hel || hel == tmp.size()) return;
  if (hel > tmp.size() - hel) {
    for (auto A : tmp)
      if (v[A][1].second <= n) ans.push_back(v[A][1].second);
  } else {
    for (auto A : tmp)
      if (v[A][1].second > n) ans.push_back(v[A][1].second - n);
  }
}
void solve() {
  cin >> n;
  for (int y = 1; y <= n; y++) cnt[y] = 0;
  for (int y = 1; y <= n; y++) cin >> a[y];
  for (int y = 1; y <= n; y++) cin >> b[y];
  for (int y = 1; y <= 2 * n; y++) {
    if (y <= n) {
      if (a[y] > n || ++cnt[a[y]] > 2) {
        cout << -1;
        return;
      }
    } else {
      if (b[y - n] > n || ++cnt[b[y - n]] > 2) {
        cout << -1;
        return;
      }
    }
  }
  for (int y = 1; y <= n; y++) v[y].clear(), mark[y] = 0;
  ans.clear();
  for (int y = 1; y <= n; y++)
    v[a[y]].push_back({b[y], y}), v[b[y]].push_back({a[y], y + n});
  for (int y = 1; y <= n; y++) {
    if (!mark[y]) tmp.clear(), dfs(y), solve2();
  }
  cout << ans.size();
  if (ans.size()) cout << '\n';
  for (auto A : ans) cout << A << ' ';
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> q;
  while (q--) solve(), cout << '\n';
  return 0;
}
