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, a[2][maxn];
class Graph {
 private:
  int n;
  vector<vector<pair<int, int>>> adj;
  int colorSet[2];
  bool dfs(int u, int color, vector<int>& c) {
    if (c[u] != -1) return c[u] == colorSet[color];
    c[u] = colorSet[color];
    for (auto& p : adj[u]) {
      int v = p.first, type = p.second;
      int vColor = type == 1 ? color : 1 - color;
      if (!dfs(v, vColor, c)) return false;
    }
    return true;
  }

 public:
  Graph(int n) : n(n) { adj.resize(n); }
  void addEdge(int u, int v, int t) {
    adj[u].push_back({v, t});
    adj[v].push_back({u, t});
  }
  vector<int> color() {
    vector<int> c(n, -1);
    for (int j = int(0); j < int(2); ++j) colorSet[j] = j;
    for (int i = int(0); i < int(n); ++i)
      if (c[i] == -1) {
        if (!dfs(i, 0, c)) return vector<int>();
        for (int j = int(0); j < int(2); ++j) colorSet[j] += 2;
      }
    return c;
  }
};
void solve() {
  vector<vector<pair<int, int>>> col(n + 1);
  for (int i = int(0); i < int(2); ++i)
    for (int j = int(0); j < int(n); ++j) col[a[i][j]].push_back({i, j});
  Graph g(n);
  for (int value = int(1); value < int(n + 1); ++value) {
    if (col[value].size() != 2) {
      cout << "-1\n";
      return;
    }
    pair<int, int> p1 = col[value][0], p2 = col[value][1];
    if (p1.second == p2.second) continue;
    if (p1.first == p2.first)
      g.addEdge(p1.second, p2.second, 2);
    else
      g.addEdge(p1.second, p2.second, 1);
  }
  vector<int> c = g.color();
  if (c.empty()) {
    cout << "-1\n";
    return;
  } else {
    map<int, int> cnt;
    for (int i = int(0); i < int(n); ++i) ++cnt[c[i]];
    auto it = cnt.end();
    int maxColor = (--it)->first;
    set<int> picked;
    int res = 0;
    for (int i = 0; i <= maxColor; i += 2) {
      res += min(cnt[i], cnt[i + 1]);
      int pick = (cnt[i] < cnt[i + 1]) ? i : i + 1;
      picked.insert(pick);
    }
    cout << res << endl;
    for (int i = int(0); i < int(n); ++i)
      if (picked.count(c[i]) > 0) cout << i + 1 << " ";
    cout << endl;
  }
}
int main() {
  int tests;
  cin >> tests;
  while (tests--) {
    cin >> n;
    for (int i = int(0); i < int(2); ++i)
      for (int j = int(0); j < int(n); ++j) cin >> a[i][j];
    solve();
  }
  return 0;
}
