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;
struct UnionFind {
  int size;
  vector<int> parent;
  UnionFind() {}
  UnionFind(int size) {
    this->size = size;
    parent.resize(size + 1);
    init();
  }
  void init() {
    for (int i = 0; i <= size; i++) parent[i] = i;
  }
  int root(int i) {
    if (parent[i] == i) return i;
    return parent[i] = root(parent[i]);
  }
  bool same(int i, int j) { return root(i) == root(j); }
  void unite(int i, int j) {
    int root_i = root(i), root_j = root(j);
    if (root_i == root_j) return;
    parent[root_i] = root_j;
  }
};
long long T;
long long n;
long long a[2][200005];
vector<pair<long long, long long> > vec[200005];
vector<long long> G[400005];
bool used[400005];
vector<long long> pvec, nvec;
void dfs(int v) {
  if (v > n) {
    used[v - n] = true;
    nvec.push_back(v);
  } else {
    used[v] = true;
    pvec.push_back(v);
  }
  for (int i = 0; i < G[v].size(); i++) {
    int u = G[v][i];
    if (u > n && used[u - n]) continue;
    if (u <= n && used[u]) continue;
    dfs(u);
  }
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  for (int z = 0; z < T; z++) {
    cin >> n;
    for (int i = 1; i <= n; i++) vec[i].clear();
    for (int i = 0; i < 2; i++) {
      for (int j = 1; j <= n; j++) {
        cin >> a[i][j];
        vec[a[i][j]].push_back(pair<long long, long long>(i, j));
      }
    }
    bool flag = true;
    for (int i = 1; i <= n; i++) {
      if (vec[i].size() != 2) flag = false;
    }
    if (!flag) {
      cout << -1 << endl;
      continue;
    }
    UnionFind uf(2 * n + 5);
    for (int i = 1; i <= 2 * n; i++) G[i].clear();
    for (int i = 1; i <= n; i++) {
      pair<long long, long long> p = vec[i][0], q = vec[i][1];
      if (p.second == q.second) continue;
      for (int j = 0; j < 2; j++) {
        long long x = abs(p.first - j), y = abs(q.first - (1 - j));
        long long u = x * n + p.second, v = y * n + q.second;
        G[u].push_back(v);
        G[v].push_back(u);
        uf.unite(u, v);
      }
    }
    flag = true;
    for (int i = 1; i <= n; i++) {
      if (uf.same(i, n + i)) flag = false;
    }
    if (!flag) {
      cout << -1 << endl;
      continue;
    }
    vector<long long> ans;
    for (int i = 1; i <= n; i++) used[i] = false;
    for (int i = 1; i <= n; i++) {
      if (a[0][i] == a[1][i]) continue;
      if (used[i]) continue;
      pvec.clear(), nvec.clear();
      dfs(i);
      if (pvec.size() > nvec.size()) swap(pvec, nvec);
      for (int i = 0; i < pvec.size(); i++) {
        if (pvec[i] > n) pvec[i] -= n;
      }
      ans.insert(ans.end(), pvec.begin(), pvec.end());
    }
    cout << ans.size() << endl;
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
    cout << endl;
  }
  return 0;
}
